100 Integral Problems and Solutions (part 4)
Hallo Guys
Assalamu'alaikum Wr.Wb.
Nama saya Hasbiansyah Cahyadi
Disini saya akan membahas soal² integral,, Mari simak pembahasannya
Berikut adalah 100 soal integral yang akan saya bahas
Pembahasan ini akan saya jadikan 4 bagian, untuk soal ini adalah problem no 76-100
Ini adalah pembahasan part ke 4, yaitu soal no 76-100
76) $\int\frac{1}{x^{2/3}(1+x^{2/3})}\ dx$
Jawab : $\int\frac{1}{x^{2/3}(1+x^{2/3})}\ dx$
Misalkan : $u=x^{1/3}$
$du=\frac{1}{3x^{2/3}}$
=$\int\frac{3}{(1+u^2)}\ du$
=$3arctan\ u+C$
=$3arctan\ x^{1/3}+C$
77) $\int\frac{sin\ x}{sin\ 2x}\ dx$
Jawab : $\int\frac{sin\ x}{sin\ 2x}\ dx$
=$\int\frac{sin\ x}{2sin\ x.cos\ x}\ dx$
=$\int\frac{sec\ x}{2}\ dx$
=$\frac{ln|tan\ x+sec\ x|}{2}+C$
78) $\int\sqrt{1+cos\ t}\ dt$
Jawab : $\int\sqrt{1+cos\ t}\ dt$
=$\int\sqrt{2cos^2\ \frac{t}{2}}\ dt$
=$\sqrt{2}\int cos\ \frac{t}{2}\ dt$
=$2\sqrt{2}sin\ \frac{t}{2}+C$
79) $\int\sqrt{1+sin\ t}\ dt$
Jawab : $\int\sqrt{1+sin\ t}\ dt$
=$\int\sqrt{1+cos\ (\frac{\pi}{2}-t)}\ dt$
=$\int\sqrt{2cos^2\ \frac{(\frac{\pi}{2}-t)}{2}}\ dt$
=$\sqrt{2}\int cos\ (\frac{\pi}{4}-\frac{t}{2})\ dt$
=$-2\sqrt{2}sin\ (\frac{\pi}{4}-\frac{t}{2})+C$
=$-2\sqrt{2}(\frac{1}{\sqrt{2}}cos\ \frac{t}{2}-\frac{1}{\sqrt{2}}sin\ \frac{t}{2})+C$
=$2sin\ \frac{t}{2}-2cos\ \frac{t}{2}+C$
80) $\int \frac{sec^2\ t}{1-tan^2\ t}\ dt$
Jawab : $\int \frac{sec^2\ t}{1-tan^2\ t}\ dt$
Misalkan : $u=tan\ t$
$du=sec^2\ t\ dt$
=$\int \frac{1}{1-u^2}\ du$
Substitusi $u=sin\ y$
$du=cos\ y\ dy$
=$\int \frac{1}{cos\ y}\ dy$
=$\int sec\ y\ dy$
=$ln|tan\ y+sec\ y|+C$
=$ln|\frac{u}{\sqrt{1-u^2}}+\frac{1}{\sqrt{1-u^2}}|+C$
=$ln|\frac{u+1}{\sqrt{1-u^2}}|+C$
=$ln|\frac{1}{\sqrt{1-u}}|+C$
=$ln|\frac{1}{\sqrt{1-tan\ t}}|+C$
81) $\int ln(x^2+x+1)\ dx$
Jawab : $\int ln(x^2+x+1)\ dx$
Misalkan $u=ln(x^2+x+1)$
$du=\frac{2x+1}{x^2+x+1}\ dx$
$dv=dx$
$v=x$
Dengan menggunakan integral parsial
=$uv-\int v\ du$
=$ln(x^2+x+1)x-\int x.\frac{2x+1}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-\int \frac{2x^2+x}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-\int 2-\frac{x+2}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x+\int\frac{x+2}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x$ $+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}+\frac{3}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{1}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{1}{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}\ dx$
Substitusi : $x+\frac{1}{2}=\frac{\sqrt{3}}{2}tan\ z$
$dx=\frac{\sqrt{3}}{2}sec^2\ z\ dz$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{sec^2\ z}{(\frac{\sqrt{3}}{2})^2sec^2\ z}\ dz$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{4}{3}\ dz$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+2z+C$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+2arctan\ \left(\frac{2x+1}{\sqrt{3}}\right)+C$
82) $\int e^{x}sin^{-1}(e^{x})\ dx$
Jawab : $\int e^{x}sin^{-1}(e^{x})\ dx$
Misalkan : $u=e^{x}$
$du=e^{x}\ dx$
$\int sin^{-1}(u)\ du$
Substitusi $u=sin\ t$
$du=cos\ t\ dt$
$\int t.cos\ t\ dt$
Dengan menggunakan integral parsial
=$t.sin\ t-\int sin\ t\ dt$
=$t.sin\ t+cos\ t+C$
=$sin^{-1}u.u+\frac{1-u^2}+C$
=$sin^{-1}e^x.e^x+\frac{1-e^{2x}}+C$
83) $\int\frac{arctan\ x}{x^2}\ dx$
Jawab : $\int\frac{arctan\ x}{x^2}\ dx$
Misalkan : $u=arctan\ x$
$du=\frac{1}{1+x^2}\ dx$
$dv=\frac{1}{x^2}\ dx$
$v=-\frac{1}{x}$
Dengan menggunakan integral parsial
=$uv-\int v\ du$
=$arctan\ x.(-\frac{1}{x})-\int -\frac{1}{x}.\frac{1}{1+x^2}\ dx$
=$-\frac{arctan\ x}{x}+\int \frac{1}{x}.\frac{1}{1+x^2}\ dx$
=$-\frac{arctan\ x}{x}+\int \frac{1}{x}-\frac{x}{1+x^2}\ dx$
=$-\frac{arctan\ x}{x}+ln|x|-\frac{1}{2}ln|1+x^2|+C$
84) $\int \frac{x^2}{\sqrt{x^2-25}}\ dx$
Jawab : $\int \frac{x^2}{\sqrt{x^2-25}}\ dx$
Substitusi : $x=5sec\ y$
$dx=5sec\ y.tan\ y\ dy$
=$\int \frac{25sec^2\ y}{\sqrt{25(sec^2\ y-1)}}5sec\ y.tan\ y\ dy$
=$\int 25sec^3\ y\ dy$
Gunakan formula
$\int sec^n\ x\ dx=\frac{sec^{n-2}\ x.tan\ x}{n-1}$ $+\frac{n-2}{n-1}\int sec^{n-2}\ x\ dx$
=$25\left(\frac{sec\ y.tan\ y}{2}+\frac{1}{2}\int sec\ y\ dy\right)$
=$\frac{25sec\ y.tan\ y}{2}+\frac{25ln|sec\ y+tan\ y|}{2}+C$
=$\frac{25.\frac{x}{5}.\frac{\sqrt{x^2-25}}{5}}{2}+\frac{25ln|\frac{x}{5}+\frac{\sqrt{x^2-25}}{5}|}{2}+C$
=$\frac{x\sqrt{x^2-25}}{2}+\frac{25ln|x+\sqrt{x^2-25}|}{2}+C$
85) $\int \frac{x^3}{(x^2+1)^2}\ dx$
Jawab : $\int \frac{x^3}{(x^2+1)^2}\ dx$
Misalkan : $u=x^2+1$
$du=2x\ dx$
=$\int \frac{u-1}{2u^2}\ du$
=$\frac{1}{2}\int\frac{1}{u}-\frac{1}{u^2}\ du$
=$\frac{1}{2}\left(ln|u|+\frac{1}{u}\right)+C$
=$\frac{1}{2}\left(ln|x^2+1|+\frac{1}{x^2+1}\right)+C$
86) $\int\frac{1}{x\sqrt{6x-x^2}}\ dx$
Jawab : $\int\frac{1}{x\sqrt{6x-x^2}}\ dx$
=$\int\frac{1}{x\sqrt{9-(x-3)^2}}\ dx$
Misalkan : $x-3=3sin\ u$
$cos\ u=\frac{\sqrt{6x-x^2}}{3}$
$cos\ \frac{u}{2}=\sqrt{\frac{\sqrt{6x-x^2}+3}{6}}$
$sin\ \frac{u}{2}=\sqrt{\frac{3-\sqrt{6x-x^2}}{6}}$
$tan\ \frac{u}{2}=\sqrt{\frac{3-\sqrt{6x-x^2}}{3+\sqrt{6x-x^2}}}$
$tan\ \frac{u}{2}=\frac{3-\sqrt{6x-x^2}}{\sqrt{9-6x+x^2}}$
$tan\ \frac{u}{2}=\frac{3-\sqrt{6x-x^2}}{x-3}$
$dx=3cos\ u\ du$
=$\int\frac{1}{(3sin\ u+3)\sqrt{9-9sin^2\ u}}3cos\ u\ du$
=$\int\frac{1}{3sin\ u+3}\ du$
=$\frac{1}{3}\int\frac{1}{sin\ u+1}\ du$
=$\frac{1}{3}\int\frac{1}{2sin\ \frac{u}{2}cos\ \frac{u}{2}+1}\ du$
=$\frac{1}{3}\int\frac{sec^2\ \frac{u}{2}}{2tan\ \frac{u}{2}+sec^2\ \frac{u}{2}}\ du$
=$\frac{1}{3}\int\frac{sec^2\ \frac{u}{2}}{(tan\ \frac{u}{2}+1)^2}\ du$
Misalkan : $y=tan\ \frac{u}{2}$
$dy=\frac{1}{2}sec^2\ \frac{u}{2}$
=$\frac{1}{3}\int\frac{2}{(y+1)^2}\ dy$
=$-\frac{2}{3(y+1)}+C$
=$-\frac{2}{3}.\frac{1}{tan\ \frac{u}{2}+1}+C$
=$-\frac{2}{3}.\frac{1}{\frac{3-\sqrt{6x-x^2}}{x-3}+1}+C$
=$-\frac{2}{3}.\frac{x-3}{x-\sqrt{6x-x^2}}+C$
=$-\frac{2}{3}.\frac{(x-3)(x+\sqrt{6x-x^2})}{2x^2-6x}+C$
=$-\frac{2}{3}.\frac{x+\sqrt{6x-x^2}}{2x}+C$
=$-\frac{x+\sqrt{6x-x^2}}{3x}+C$
=$-\frac{\sqrt{6x-x^2}}{3x}+C$
87) $\int\frac{3x+2}{(x^2+4)^{3/2}}\ dx$
Jawab : $\int\frac{3x+2}{(x^2+4)^{3/2}}\ dx$
Substitusi $x=2tan\ u$
$dx=2sec^2\ u$
=$\int\frac{6tan\ u+2}{(4tan^2\ u+4)^{3/2}}2sec^2\ u\ du$
=$\int\frac{6tan\ u+2}{(2sec\ u)^3}2sec^2\ u\ du$
=$\int\frac{3tan\ u+1}{2sec\ u}\ du$
=$\int\frac{3sin\ u+cos\ u}{2}\ du$
=$\frac{1}{2}\left(-3cos\ u+sin\ u\right)+C$
=$\frac{1}{2}\left(\frac{-6}{\sqrt{x^2+4}}+\frac{x}{\sqrt{x^2+4}}\right)+C$
=$\frac{x-6}{2\sqrt{x^2+4}}+C$
88) $\int x^{3/2}ln\ x\ dx$
Jawab : $\int x^{3/2}ln\ x\ dx$
Misalkan : $u=ln\ x$
$du=\frac{1}{x}\ dx$
=$\int e^{5u/2}u\ dx$
Dengan menggunakan parsial didapat
=$u.\frac{2}{5}e^{5u/2}-\int e^{5u/2}\ du$
=$\frac{2u}{5}e^{5u/2}-\frac{2}{5}e^{5u/2}+C$
=$\frac{2e^{5ln\ x/2}}{5}(ln\ x-1)+C$
=$\frac{2x^{5/2}}{5}(ln\ x-1)+C$
89) $\int \frac{1+sin^2\ x}{sec\ x.csc\ x}\ dx$
Jawab : $\int \frac{1+sin^2\ x}{sec\ x.csc\ x}\ dx$
=$\int (sin\ x+sin^3\ x)cos\ x\ dx$
Misalkan : $u=sin\ x$
$du=cos\ x\ dx$
=$\int (u+u^3)\ du$
=$\frac{1}{2}u^2+\frac{1}{4}u^4+C$
=$\frac{1}{2}sin^2\ x+\frac{1}{4}sin^4\ x+C$
90) $\int\frac{exp(\sqrt{sin\ x})}{(sec\ x)\sqrt{sin\ x}}\ dx$
Jawab : $\int\frac{exp(\sqrt{sin\ x})}{(sec\ x)\sqrt{sin\ x}}\ dx$
=$\int\frac{e^{\sqrt{sin\ x}}cos\ x}{\sqrt{sin\ x}}\ dx$
Misalkan : $y=sin\ x$
$dy=cos\ x\ dx$
=$\int\frac{e^{\sqrt{y}}}{\sqrt{y}}\ dy$
=$2e^{\sqrt{y}}+C$
=$2e^{\sqrt{sin\ x}}+C$
91) $\int xe^{x}sin\ x\ dx$
Jawab : $\int xe^{x}sin\ x\ dx$
Misalkan : $u=x$
$du=dx$
$dv=e^{x}sin\ x\ dx$
$v=\frac{e^{x}sin\ x-e^{x}cos\ x}{2}$
Dengan integral parsial sehingga
=$x\frac{e^{x}sin\ x-e^{x}cos\ x}{2}-\int \frac{e^{x}sin\ x-e^{x}cos\ x}{2}\ dx$
=$x\frac{e^{x}sin\ x-e^{x}cos\ x}{2}-\frac{e^{x}sin\ x-e^{x}cos\ x}{4}$ $+\frac{e^{x}sin\ x+e^{x}cos\ x}{4}+C$
92) $\int x^2.exp(x^{3/2})\ dx$
Jawab : $\int x^2.exp(x^{3/2})\ dx$
=$\int x^2.e^{x^{3/2}}\ dx$
Misalkan : $u=x^{3/2}$
$du=\frac{3}{2}x^{1/2}\ dx$
=$\int \frac{2}{3}u.e^{u}\ du$
=$\frac{2}{3}e^{u}(u-1)+C$
=$\frac{2}{3}e^{x^{3/2}}(x^{3/2}-1)+C$
93) $\int\frac{arctan\ x}{(x-1)^3}\ dx$
Jawab : $\int\frac{arctan\ x}{(x-1)^3}\ dx$
Misalkan : $u=arctan\ x$
$x=tan\ u$
$dx=sec^2\ u\ du$
=$\int\frac{u.sec^2\ u}{(tan\ u-1)^3}\ du$
Misalkan : $y=u$
$dy=du$
$dz=\frac{sec^2\ u}{(tan\ u-1)^3}\ du$
$z=-\frac{1}{2(tan\ u-1)^2}$
Dengan parsial diperoleh
=$-\frac{u}{2(tan\ u-1)^2}-\int -\frac{1}{2(tan\ u-1)^2}\ du$
=$-\frac{u}{2(tan\ u-1)^2}+\frac{1}{2}\int \frac{sec^2\ u}{(tan\ u-1)^2(tan^2\ u+1)}\ du$
Misalkan : $v=tan\ u$
$dv=sec^2\ u\ du$
=$-\frac{u}{2(tan\ u-1)^2}+\int \frac{1}{2(v-1)^2(v^2+1)}\ dv$
=$-\frac{u}{2(tan\ u-1)^2}$ $+\int \frac{-1}{4(v-1)}+\frac{1}{4(v-1)^2}+\frac{v}{4(v^2+1)}\ dv$
=$-\frac{u}{2(tan\ u-1)^2}+\frac{-ln|v-1|-\frac{1}{v-1}+\frac{1}{2}ln|v^2+1|}{4}+C$
=$-\frac{u}{2(tan\ u-1)^2}$ $+\frac{-ln|tan\ u-1|-\frac{1}{tan\ u-1}+\frac{1}{2}ln|tan^2\ u+1|}{4}+C$
=$-\frac{arctan\ x}{2(x-1)^2}+\frac{-ln|x-1|-\frac{1}{x-1}+\frac{1}{2}ln|x^2+1|}{4}+C$
94) $\int ln(1+\sqrt{x})\ dx$
Jawab : $\int ln(1+\sqrt{x})\ dx$
Misalkan : $1+\sqrt{x}=y$
$x=(y-1)^2$
$dx=2(y-1)\ dy$
=$\int ln(y).2(y-1)\ dy$
Misalkan : $u=ln(y)$
$du=\frac{1}{y}\ dy$
$dv=2(y-1)\ dy$
$v=(y-1)^2$
Dengan parsial diperoleh
=$ln(y).(y-1)^2-\int \frac{(y-1)^2}{y}\ dy$
=$ln(y).(y-1)^2-\int \frac{y^2-2y+1}{y}\ dy$
=$ln(y).(y-1)^2-\int y-2+\frac{1}{y}\ dy$
=$ln(y).(y-1)^2-\frac{1}{2}y^2$ $+2y-ln(y)+C$
=$ln(1+\sqrt{x}).x-\frac{1}{2}(1+\sqrt{x})^2$ $+2(1+\sqrt{x})-ln(1+\sqrt{x})+C$
=$ln(1+\sqrt{x}).x-\sqrt{x}-\frac{1}{2}x+2\sqrt{x}$ $-ln(1+\sqrt{x})+C$
=$ln(1+\sqrt{x}).x+\sqrt{x}-\frac{1}{2}x$ $-ln(1+\sqrt{x})+C$
95) $\int\frac{2x+3}{\sqrt{3+6x-9x^2}}\ dx$
Jawab : $\int\frac{2x+3}{\sqrt{3+6x-9x^2}}\ dx$
=$\frac{1}{\sqrt{3}}\int\frac{(-6x+2)-11}{-3\sqrt{1+2x-3x^2}}\ dx$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx$ $+\frac{11}{3\sqrt{3}}\int\frac{1}{\sqrt{1+2x-3x^2}}\ dx$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx$ $+\frac{11}{9}\int\frac{1}{\sqrt{(\frac{2}{3})^2-(x-\frac{1}{3})^2}}\ dx$
Substitusi : $x-\frac{1}{3}=\frac{2}{3}sin\ u$
$dx=\frac{2}{3}cos\ u\ du$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx$ $+\frac{11}{9}\int\frac{2cos\ u}{3\sqrt{(\frac{2}{3})^2(1-sin^2\ u)}}\ du$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx+\frac{11}{9}\int\frac{2cos\ u}{2cos\ u}\ du$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx+\frac{11}{9}u+C$
=$-\frac{2\sqrt{1+2x-3x^2}}{3\sqrt{3}}+\frac{11}{9}arcsin\ \left(\frac{3x-1}{2}\right)+C$
96) $\int\frac{1}{2+2sin\ \theta+cos\ \theta}\ d\theta$
Jawab : $\int\frac{1}{2+2sin\ \theta+cos\ \theta}\ d\theta$
=$\int\frac{1}{1+sin^2\ \frac{\theta}{2}+cos^2\ \frac{\theta}{2}+2sin\ \theta+cos^2\ \frac{\theta}{2}-sin^2\ \frac{\theta}{2}}\ d\theta$
Assalamu'alaikum Wr.Wb.
Nama saya Hasbiansyah Cahyadi
Disini saya akan membahas soal² integral,, Mari simak pembahasannya
Berikut adalah 100 soal integral yang akan saya bahas
Pembahasan ini akan saya jadikan 4 bagian, untuk soal ini adalah problem no 76-100
 |
| Soal Integral Aljabar dan Trigonometri |
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| Soal Integral Aljabar dan Trigonometri |
Ini adalah pembahasan part ke 4, yaitu soal no 76-100
76) $\int\frac{1}{x^{2/3}(1+x^{2/3})}\ dx$
Jawab : $\int\frac{1}{x^{2/3}(1+x^{2/3})}\ dx$
Misalkan : $u=x^{1/3}$
$du=\frac{1}{3x^{2/3}}$
=$\int\frac{3}{(1+u^2)}\ du$
=$3arctan\ u+C$
=$3arctan\ x^{1/3}+C$
77) $\int\frac{sin\ x}{sin\ 2x}\ dx$
Jawab : $\int\frac{sin\ x}{sin\ 2x}\ dx$
=$\int\frac{sin\ x}{2sin\ x.cos\ x}\ dx$
=$\int\frac{sec\ x}{2}\ dx$
=$\frac{ln|tan\ x+sec\ x|}{2}+C$
78) $\int\sqrt{1+cos\ t}\ dt$
Jawab : $\int\sqrt{1+cos\ t}\ dt$
=$\int\sqrt{2cos^2\ \frac{t}{2}}\ dt$
=$\sqrt{2}\int cos\ \frac{t}{2}\ dt$
=$2\sqrt{2}sin\ \frac{t}{2}+C$
79) $\int\sqrt{1+sin\ t}\ dt$
Jawab : $\int\sqrt{1+sin\ t}\ dt$
=$\int\sqrt{1+cos\ (\frac{\pi}{2}-t)}\ dt$
=$\int\sqrt{2cos^2\ \frac{(\frac{\pi}{2}-t)}{2}}\ dt$
=$\sqrt{2}\int cos\ (\frac{\pi}{4}-\frac{t}{2})\ dt$
=$-2\sqrt{2}sin\ (\frac{\pi}{4}-\frac{t}{2})+C$
=$-2\sqrt{2}(\frac{1}{\sqrt{2}}cos\ \frac{t}{2}-\frac{1}{\sqrt{2}}sin\ \frac{t}{2})+C$
=$2sin\ \frac{t}{2}-2cos\ \frac{t}{2}+C$
80) $\int \frac{sec^2\ t}{1-tan^2\ t}\ dt$
Jawab : $\int \frac{sec^2\ t}{1-tan^2\ t}\ dt$
Misalkan : $u=tan\ t$
$du=sec^2\ t\ dt$
=$\int \frac{1}{1-u^2}\ du$
Substitusi $u=sin\ y$
$du=cos\ y\ dy$
=$\int \frac{1}{cos\ y}\ dy$
=$\int sec\ y\ dy$
=$ln|tan\ y+sec\ y|+C$
=$ln|\frac{u}{\sqrt{1-u^2}}+\frac{1}{\sqrt{1-u^2}}|+C$
=$ln|\frac{u+1}{\sqrt{1-u^2}}|+C$
=$ln|\frac{1}{\sqrt{1-u}}|+C$
=$ln|\frac{1}{\sqrt{1-tan\ t}}|+C$
81) $\int ln(x^2+x+1)\ dx$
Jawab : $\int ln(x^2+x+1)\ dx$
Misalkan $u=ln(x^2+x+1)$
$du=\frac{2x+1}{x^2+x+1}\ dx$
$dv=dx$
$v=x$
Dengan menggunakan integral parsial
=$uv-\int v\ du$
=$ln(x^2+x+1)x-\int x.\frac{2x+1}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-\int \frac{2x^2+x}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-\int 2-\frac{x+2}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x+\int\frac{x+2}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x$ $+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}+\frac{3}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{1}{x^2+x+1}\ dx$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{1}{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}\ dx$
Substitusi : $x+\frac{1}{2}=\frac{\sqrt{3}}{2}tan\ z$
$dx=\frac{\sqrt{3}}{2}sec^2\ z\ dz$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{sec^2\ z}{(\frac{\sqrt{3}}{2})^2sec^2\ z}\ dz$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+\frac{3}{2}\int\frac{4}{3}\ dz$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+2z+C$
=$ln(x^2+x+1)x-2x+\frac{1}{2}ln(x^2+x+1)$ $+2arctan\ \left(\frac{2x+1}{\sqrt{3}}\right)+C$
82) $\int e^{x}sin^{-1}(e^{x})\ dx$
Jawab : $\int e^{x}sin^{-1}(e^{x})\ dx$
Misalkan : $u=e^{x}$
$du=e^{x}\ dx$
$\int sin^{-1}(u)\ du$
Substitusi $u=sin\ t$
$du=cos\ t\ dt$
$\int t.cos\ t\ dt$
Dengan menggunakan integral parsial
=$t.sin\ t-\int sin\ t\ dt$
=$t.sin\ t+cos\ t+C$
=$sin^{-1}u.u+\frac{1-u^2}+C$
=$sin^{-1}e^x.e^x+\frac{1-e^{2x}}+C$
83) $\int\frac{arctan\ x}{x^2}\ dx$
Jawab : $\int\frac{arctan\ x}{x^2}\ dx$
Misalkan : $u=arctan\ x$
$du=\frac{1}{1+x^2}\ dx$
$dv=\frac{1}{x^2}\ dx$
$v=-\frac{1}{x}$
Dengan menggunakan integral parsial
=$uv-\int v\ du$
=$arctan\ x.(-\frac{1}{x})-\int -\frac{1}{x}.\frac{1}{1+x^2}\ dx$
=$-\frac{arctan\ x}{x}+\int \frac{1}{x}.\frac{1}{1+x^2}\ dx$
=$-\frac{arctan\ x}{x}+\int \frac{1}{x}-\frac{x}{1+x^2}\ dx$
=$-\frac{arctan\ x}{x}+ln|x|-\frac{1}{2}ln|1+x^2|+C$
84) $\int \frac{x^2}{\sqrt{x^2-25}}\ dx$
Jawab : $\int \frac{x^2}{\sqrt{x^2-25}}\ dx$
Substitusi : $x=5sec\ y$
$dx=5sec\ y.tan\ y\ dy$
=$\int \frac{25sec^2\ y}{\sqrt{25(sec^2\ y-1)}}5sec\ y.tan\ y\ dy$
=$\int 25sec^3\ y\ dy$
Gunakan formula
$\int sec^n\ x\ dx=\frac{sec^{n-2}\ x.tan\ x}{n-1}$ $+\frac{n-2}{n-1}\int sec^{n-2}\ x\ dx$
=$25\left(\frac{sec\ y.tan\ y}{2}+\frac{1}{2}\int sec\ y\ dy\right)$
=$\frac{25sec\ y.tan\ y}{2}+\frac{25ln|sec\ y+tan\ y|}{2}+C$
=$\frac{25.\frac{x}{5}.\frac{\sqrt{x^2-25}}{5}}{2}+\frac{25ln|\frac{x}{5}+\frac{\sqrt{x^2-25}}{5}|}{2}+C$
=$\frac{x\sqrt{x^2-25}}{2}+\frac{25ln|x+\sqrt{x^2-25}|}{2}+C$
85) $\int \frac{x^3}{(x^2+1)^2}\ dx$
Jawab : $\int \frac{x^3}{(x^2+1)^2}\ dx$
Misalkan : $u=x^2+1$
$du=2x\ dx$
=$\int \frac{u-1}{2u^2}\ du$
=$\frac{1}{2}\int\frac{1}{u}-\frac{1}{u^2}\ du$
=$\frac{1}{2}\left(ln|u|+\frac{1}{u}\right)+C$
=$\frac{1}{2}\left(ln|x^2+1|+\frac{1}{x^2+1}\right)+C$
86) $\int\frac{1}{x\sqrt{6x-x^2}}\ dx$
Jawab : $\int\frac{1}{x\sqrt{6x-x^2}}\ dx$
=$\int\frac{1}{x\sqrt{9-(x-3)^2}}\ dx$
Misalkan : $x-3=3sin\ u$
$cos\ u=\frac{\sqrt{6x-x^2}}{3}$
$cos\ \frac{u}{2}=\sqrt{\frac{\sqrt{6x-x^2}+3}{6}}$
$sin\ \frac{u}{2}=\sqrt{\frac{3-\sqrt{6x-x^2}}{6}}$
$tan\ \frac{u}{2}=\sqrt{\frac{3-\sqrt{6x-x^2}}{3+\sqrt{6x-x^2}}}$
$tan\ \frac{u}{2}=\frac{3-\sqrt{6x-x^2}}{\sqrt{9-6x+x^2}}$
$tan\ \frac{u}{2}=\frac{3-\sqrt{6x-x^2}}{x-3}$
$dx=3cos\ u\ du$
=$\int\frac{1}{(3sin\ u+3)\sqrt{9-9sin^2\ u}}3cos\ u\ du$
=$\int\frac{1}{3sin\ u+3}\ du$
=$\frac{1}{3}\int\frac{1}{sin\ u+1}\ du$
=$\frac{1}{3}\int\frac{1}{2sin\ \frac{u}{2}cos\ \frac{u}{2}+1}\ du$
=$\frac{1}{3}\int\frac{sec^2\ \frac{u}{2}}{2tan\ \frac{u}{2}+sec^2\ \frac{u}{2}}\ du$
=$\frac{1}{3}\int\frac{sec^2\ \frac{u}{2}}{(tan\ \frac{u}{2}+1)^2}\ du$
Misalkan : $y=tan\ \frac{u}{2}$
$dy=\frac{1}{2}sec^2\ \frac{u}{2}$
=$\frac{1}{3}\int\frac{2}{(y+1)^2}\ dy$
=$-\frac{2}{3(y+1)}+C$
=$-\frac{2}{3}.\frac{1}{tan\ \frac{u}{2}+1}+C$
=$-\frac{2}{3}.\frac{1}{\frac{3-\sqrt{6x-x^2}}{x-3}+1}+C$
=$-\frac{2}{3}.\frac{x-3}{x-\sqrt{6x-x^2}}+C$
=$-\frac{2}{3}.\frac{(x-3)(x+\sqrt{6x-x^2})}{2x^2-6x}+C$
=$-\frac{2}{3}.\frac{x+\sqrt{6x-x^2}}{2x}+C$
=$-\frac{x+\sqrt{6x-x^2}}{3x}+C$
=$-\frac{\sqrt{6x-x^2}}{3x}+C$
87) $\int\frac{3x+2}{(x^2+4)^{3/2}}\ dx$
Jawab : $\int\frac{3x+2}{(x^2+4)^{3/2}}\ dx$
Substitusi $x=2tan\ u$
$dx=2sec^2\ u$
=$\int\frac{6tan\ u+2}{(4tan^2\ u+4)^{3/2}}2sec^2\ u\ du$
=$\int\frac{6tan\ u+2}{(2sec\ u)^3}2sec^2\ u\ du$
=$\int\frac{3tan\ u+1}{2sec\ u}\ du$
=$\int\frac{3sin\ u+cos\ u}{2}\ du$
=$\frac{1}{2}\left(-3cos\ u+sin\ u\right)+C$
=$\frac{1}{2}\left(\frac{-6}{\sqrt{x^2+4}}+\frac{x}{\sqrt{x^2+4}}\right)+C$
=$\frac{x-6}{2\sqrt{x^2+4}}+C$
88) $\int x^{3/2}ln\ x\ dx$
Jawab : $\int x^{3/2}ln\ x\ dx$
Misalkan : $u=ln\ x$
$du=\frac{1}{x}\ dx$
=$\int e^{5u/2}u\ dx$
Dengan menggunakan parsial didapat
=$u.\frac{2}{5}e^{5u/2}-\int e^{5u/2}\ du$
=$\frac{2u}{5}e^{5u/2}-\frac{2}{5}e^{5u/2}+C$
=$\frac{2e^{5ln\ x/2}}{5}(ln\ x-1)+C$
=$\frac{2x^{5/2}}{5}(ln\ x-1)+C$
89) $\int \frac{1+sin^2\ x}{sec\ x.csc\ x}\ dx$
Jawab : $\int \frac{1+sin^2\ x}{sec\ x.csc\ x}\ dx$
=$\int (sin\ x+sin^3\ x)cos\ x\ dx$
Misalkan : $u=sin\ x$
$du=cos\ x\ dx$
=$\int (u+u^3)\ du$
=$\frac{1}{2}u^2+\frac{1}{4}u^4+C$
=$\frac{1}{2}sin^2\ x+\frac{1}{4}sin^4\ x+C$
90) $\int\frac{exp(\sqrt{sin\ x})}{(sec\ x)\sqrt{sin\ x}}\ dx$
Jawab : $\int\frac{exp(\sqrt{sin\ x})}{(sec\ x)\sqrt{sin\ x}}\ dx$
=$\int\frac{e^{\sqrt{sin\ x}}cos\ x}{\sqrt{sin\ x}}\ dx$
Misalkan : $y=sin\ x$
$dy=cos\ x\ dx$
=$\int\frac{e^{\sqrt{y}}}{\sqrt{y}}\ dy$
=$2e^{\sqrt{y}}+C$
=$2e^{\sqrt{sin\ x}}+C$
91) $\int xe^{x}sin\ x\ dx$
Jawab : $\int xe^{x}sin\ x\ dx$
Misalkan : $u=x$
$du=dx$
$dv=e^{x}sin\ x\ dx$
$v=\frac{e^{x}sin\ x-e^{x}cos\ x}{2}$
Dengan integral parsial sehingga
=$x\frac{e^{x}sin\ x-e^{x}cos\ x}{2}-\int \frac{e^{x}sin\ x-e^{x}cos\ x}{2}\ dx$
=$x\frac{e^{x}sin\ x-e^{x}cos\ x}{2}-\frac{e^{x}sin\ x-e^{x}cos\ x}{4}$ $+\frac{e^{x}sin\ x+e^{x}cos\ x}{4}+C$
92) $\int x^2.exp(x^{3/2})\ dx$
Jawab : $\int x^2.exp(x^{3/2})\ dx$
=$\int x^2.e^{x^{3/2}}\ dx$
Misalkan : $u=x^{3/2}$
$du=\frac{3}{2}x^{1/2}\ dx$
=$\int \frac{2}{3}u.e^{u}\ du$
=$\frac{2}{3}e^{u}(u-1)+C$
=$\frac{2}{3}e^{x^{3/2}}(x^{3/2}-1)+C$
93) $\int\frac{arctan\ x}{(x-1)^3}\ dx$
Jawab : $\int\frac{arctan\ x}{(x-1)^3}\ dx$
Misalkan : $u=arctan\ x$
$x=tan\ u$
$dx=sec^2\ u\ du$
=$\int\frac{u.sec^2\ u}{(tan\ u-1)^3}\ du$
Misalkan : $y=u$
$dy=du$
$dz=\frac{sec^2\ u}{(tan\ u-1)^3}\ du$
$z=-\frac{1}{2(tan\ u-1)^2}$
Dengan parsial diperoleh
=$-\frac{u}{2(tan\ u-1)^2}-\int -\frac{1}{2(tan\ u-1)^2}\ du$
=$-\frac{u}{2(tan\ u-1)^2}+\frac{1}{2}\int \frac{sec^2\ u}{(tan\ u-1)^2(tan^2\ u+1)}\ du$
Misalkan : $v=tan\ u$
$dv=sec^2\ u\ du$
=$-\frac{u}{2(tan\ u-1)^2}+\int \frac{1}{2(v-1)^2(v^2+1)}\ dv$
=$-\frac{u}{2(tan\ u-1)^2}$ $+\int \frac{-1}{4(v-1)}+\frac{1}{4(v-1)^2}+\frac{v}{4(v^2+1)}\ dv$
=$-\frac{u}{2(tan\ u-1)^2}+\frac{-ln|v-1|-\frac{1}{v-1}+\frac{1}{2}ln|v^2+1|}{4}+C$
=$-\frac{u}{2(tan\ u-1)^2}$ $+\frac{-ln|tan\ u-1|-\frac{1}{tan\ u-1}+\frac{1}{2}ln|tan^2\ u+1|}{4}+C$
=$-\frac{arctan\ x}{2(x-1)^2}+\frac{-ln|x-1|-\frac{1}{x-1}+\frac{1}{2}ln|x^2+1|}{4}+C$
94) $\int ln(1+\sqrt{x})\ dx$
Jawab : $\int ln(1+\sqrt{x})\ dx$
Misalkan : $1+\sqrt{x}=y$
$x=(y-1)^2$
$dx=2(y-1)\ dy$
=$\int ln(y).2(y-1)\ dy$
Misalkan : $u=ln(y)$
$du=\frac{1}{y}\ dy$
$dv=2(y-1)\ dy$
$v=(y-1)^2$
Dengan parsial diperoleh
=$ln(y).(y-1)^2-\int \frac{(y-1)^2}{y}\ dy$
=$ln(y).(y-1)^2-\int \frac{y^2-2y+1}{y}\ dy$
=$ln(y).(y-1)^2-\int y-2+\frac{1}{y}\ dy$
=$ln(y).(y-1)^2-\frac{1}{2}y^2$ $+2y-ln(y)+C$
=$ln(1+\sqrt{x}).x-\frac{1}{2}(1+\sqrt{x})^2$ $+2(1+\sqrt{x})-ln(1+\sqrt{x})+C$
=$ln(1+\sqrt{x}).x-\sqrt{x}-\frac{1}{2}x+2\sqrt{x}$ $-ln(1+\sqrt{x})+C$
=$ln(1+\sqrt{x}).x+\sqrt{x}-\frac{1}{2}x$ $-ln(1+\sqrt{x})+C$
95) $\int\frac{2x+3}{\sqrt{3+6x-9x^2}}\ dx$
Jawab : $\int\frac{2x+3}{\sqrt{3+6x-9x^2}}\ dx$
=$\frac{1}{\sqrt{3}}\int\frac{(-6x+2)-11}{-3\sqrt{1+2x-3x^2}}\ dx$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx$ $+\frac{11}{3\sqrt{3}}\int\frac{1}{\sqrt{1+2x-3x^2}}\ dx$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx$ $+\frac{11}{9}\int\frac{1}{\sqrt{(\frac{2}{3})^2-(x-\frac{1}{3})^2}}\ dx$
Substitusi : $x-\frac{1}{3}=\frac{2}{3}sin\ u$
$dx=\frac{2}{3}cos\ u\ du$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx$ $+\frac{11}{9}\int\frac{2cos\ u}{3\sqrt{(\frac{2}{3})^2(1-sin^2\ u)}}\ du$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx+\frac{11}{9}\int\frac{2cos\ u}{2cos\ u}\ du$
=$-\frac{1}{3\sqrt{3}}\int\frac{(-6x+2)}{\sqrt{1+2x-3x^2}}\ dx+\frac{11}{9}u+C$
=$-\frac{2\sqrt{1+2x-3x^2}}{3\sqrt{3}}+\frac{11}{9}arcsin\ \left(\frac{3x-1}{2}\right)+C$
96) $\int\frac{1}{2+2sin\ \theta+cos\ \theta}\ d\theta$
Jawab : $\int\frac{1}{2+2sin\ \theta+cos\ \theta}\ d\theta$
=$\int\frac{1}{1+sin^2\ \frac{\theta}{2}+cos^2\ \frac{\theta}{2}+2sin\ \theta+cos^2\ \frac{\theta}{2}-sin^2\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{1}{1+2cos^2\ \frac{\theta}{2}+2sin\ \theta}\ d\theta$
=$\int\frac{1}{1+2cos^2\ \frac{\theta}{2}+4sin\ \frac{\theta}{2}cos\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{sec^2\ \frac{\theta}{2}+2+4tan \frac{\theta}{2}}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{tan^2\ \frac{\theta}{2}+3+4tan \frac{\theta}{2}}\ d\theta$
Misalkan : $u=tan\ \frac{\theta}{2}$
$du=\frac{1}{2}sec^2\ \frac{\theta}{2}\ d\theta$
=$\int\frac{2}{u^2+3+4u}\ du$
=$\int\frac{2}{(u+1)(u+3)}\ du$
=$\int\frac{1}{u+1}-\frac{1}{u+3}\ du$
=$ln|u+1|-ln|u+3|+C$
=$ln\left|\frac{tan\ \frac{\theta}{2}+1}{tan\ \frac{\theta}{2}+3}\right|+C$
97) $\int\frac{sin^3\ \theta}{cos\ \theta-1}\ d\theta$
Jawab : $\int\frac{sin^3\ \theta}{cos\ \theta-1}\ d\theta$
=$\int\frac{(1-cos^2\ \theta)sin\ \theta}{cos\ \theta-1}\ d\theta$
=$\int(1+cos\ \theta)(-sin\ \theta)\ d\theta$
Misalkan : $u=cos\ \theta$
$du=-sin\ \theta\ d\theta$
=$\int 1+u\ du$
=$u+\frac{1}{2}u^2+C$
=$cos\ \theta+\frac{1}{2}cos^2\ \theta+C$
98) $\int x^{3/2}tan^{-1}\ (x^{1/2})\ dx$
Jawab : $\int x^{3/2}tan^{-1}\ (x^{1/2})\ dx$
Substitusi : $x=y^2$
$dx=2y\ dy$
=$\int 2y^4tan^{-1}\ y\ dy$
Misalkan : $u=tan^{-1}\ y$
$du=\frac{1}{y^2+1}\ dy$
$dv=2y^4\ dy$
$v=\frac{2}{5}y^5$
Dengan menggunakan integral parsial diperoleh
=$\frac{2}{5}y^5tan^{-1}\ y-\int \frac{2}{5}y^5.\frac{1}{y^2+1}\ dy$
=$\frac{2}{5}y^5tan^{-1}\ y-\frac{2}{5}\int \frac{y^5}{y^2+1}\ dy$
=$\frac{2}{5}y^5tan^{-1}\ y-\frac{2}{5}\int y^3-y+\frac{y}{y^2+1}\ dy$
=$\frac{2}{5}y^5tan^{-1}\ y-\frac{1}{10}y^4+\frac{1}{5}y^2-\frac{1}{5}\int \frac{2y}{y^2+1}\ dy$
=$\frac{2y^5tan^{-1}\ y}{5}-\frac{y^4}{10}+\frac{y^2}{5}-\frac{ln|y^2+1|}{5}+C$
=$\frac{2x^{5/2}tan^{-1}\ \sqrt{x}}{5}-\frac{x^2}{10}+\frac{x}{5}-\frac{ln|x+1|}{5}+C$
99) $\int sec^{-1}\ \sqrt{x}\ dx$
Jawab : $\int sec^{-1}\ \sqrt{x}\ dx$
Substitusi : $x=y^2$
$dx=2y\ dy$
=$\int sec^{-1}\ y.2y\ dy$
Misalkan : $u=sec^{-1}\ y$
$du=\frac{1}{y\sqrt{y^2-1}}\ dy$
$dv=2y\ dy$
$v=y^2$
Dengan menggunakan integral parsial diperoleh
=$y^2sec^{-1}\ y-\int \frac{y}{\sqrt{y^2-1}}\ dy$
Misalkan : $z=y^2-1$
$dz=2y\ dy$
=$y^2sec^{-1}\ y-\int \frac{1}{2\sqrt{z}}\ dz$
=$y^2sec^{-1}\ y-\sqrt{z}+C$
=$y^2sec^{-1}\ y-\sqrt{y^2-1}+C$
=$xsec^{-1}\ \sqrt{x}-\sqrt{x-1}+C$
100) $\int x.\left(\frac{1-x^2}{1+x^2}\right)^{1/2}\ dx$
Jawab : $\int x.\left(\frac{1-x^2}{1+x^2}\right)^{1/2}\ dx$
Misalkan : $u=(1+x^2)^{1/2}$
$du=\frac{x}{(1+x^2)^{1/2}}\ dx$
=$\int (1-(u^2-1))^{1/2}\ du$
=$\int (2-u^2)^{1/2}\ du$
Substitusi : $u=\sqrt{2}sin\ y$
$du=\sqrt{2}cos\ y\ dy$
=$\int \sqrt{2(1-sin^2\ y)}.\sqrt{2}cos\ y\ dy$
=$\int 2cos^2\ y\ dy$
=$\int cos\ 2y+1\ dy$
=$\frac{1}{2}sin\ 2y+y+C$
=$sin\ y.cos\ y+y+C$
=$\frac{u}{\sqrt{2}}.\frac{\sqrt{2-u^2}}{\sqrt{2}}+arcsin\ \frac{u}{\sqrt{2}}+C$
=$\frac{\sqrt{1+x^2}}{\sqrt{2}}.\frac{\sqrt{1-x^2}}{\sqrt{2}}+arcsin\ \frac{\sqrt{1+x^2}}{\sqrt{2}}+C$
=$\frac{\sqrt{1-x^4}}{2}+arcsin\ \sqrt{\frac{1+x^2}{2}}+C$
Alhamdulillah pembahasannya selesai 😀😁
Saya pribadi meminta maaf jika ternyata masih banyak kekurangan dalam pembahasan ini. Apabila ada kritik dan masukan tuliskan di kolom komentar. Semoga pembahasannya bisa bermanfaat buat kalian semua...
Terima kasih telah berkunjung ke blog saya....🙏
Salam math lover";💕👌
=$\int\frac{sec^2\ \frac{\theta}{2}}{sec^2\ \frac{\theta}{2}+2+4tan \frac{\theta}{2}}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{tan^2\ \frac{\theta}{2}+3+4tan \frac{\theta}{2}}\ d\theta$
Misalkan : $u=tan\ \frac{\theta}{2}$
$du=\frac{1}{2}sec^2\ \frac{\theta}{2}\ d\theta$
=$\int\frac{2}{u^2+3+4u}\ du$
=$\int\frac{2}{(u+1)(u+3)}\ du$
=$\int\frac{1}{u+1}-\frac{1}{u+3}\ du$
=$ln|u+1|-ln|u+3|+C$
=$ln\left|\frac{tan\ \frac{\theta}{2}+1}{tan\ \frac{\theta}{2}+3}\right|+C$
97) $\int\frac{sin^3\ \theta}{cos\ \theta-1}\ d\theta$
Jawab : $\int\frac{sin^3\ \theta}{cos\ \theta-1}\ d\theta$
=$\int\frac{(1-cos^2\ \theta)sin\ \theta}{cos\ \theta-1}\ d\theta$
=$\int(1+cos\ \theta)(-sin\ \theta)\ d\theta$
Misalkan : $u=cos\ \theta$
$du=-sin\ \theta\ d\theta$
=$\int 1+u\ du$
=$u+\frac{1}{2}u^2+C$
=$cos\ \theta+\frac{1}{2}cos^2\ \theta+C$
98) $\int x^{3/2}tan^{-1}\ (x^{1/2})\ dx$
Jawab : $\int x^{3/2}tan^{-1}\ (x^{1/2})\ dx$
Substitusi : $x=y^2$
$dx=2y\ dy$
=$\int 2y^4tan^{-1}\ y\ dy$
Misalkan : $u=tan^{-1}\ y$
$du=\frac{1}{y^2+1}\ dy$
$dv=2y^4\ dy$
$v=\frac{2}{5}y^5$
Dengan menggunakan integral parsial diperoleh
=$\frac{2}{5}y^5tan^{-1}\ y-\int \frac{2}{5}y^5.\frac{1}{y^2+1}\ dy$
=$\frac{2}{5}y^5tan^{-1}\ y-\frac{2}{5}\int \frac{y^5}{y^2+1}\ dy$
=$\frac{2}{5}y^5tan^{-1}\ y-\frac{2}{5}\int y^3-y+\frac{y}{y^2+1}\ dy$
=$\frac{2}{5}y^5tan^{-1}\ y-\frac{1}{10}y^4+\frac{1}{5}y^2-\frac{1}{5}\int \frac{2y}{y^2+1}\ dy$
=$\frac{2y^5tan^{-1}\ y}{5}-\frac{y^4}{10}+\frac{y^2}{5}-\frac{ln|y^2+1|}{5}+C$
=$\frac{2x^{5/2}tan^{-1}\ \sqrt{x}}{5}-\frac{x^2}{10}+\frac{x}{5}-\frac{ln|x+1|}{5}+C$
99) $\int sec^{-1}\ \sqrt{x}\ dx$
Jawab : $\int sec^{-1}\ \sqrt{x}\ dx$
Substitusi : $x=y^2$
$dx=2y\ dy$
=$\int sec^{-1}\ y.2y\ dy$
Misalkan : $u=sec^{-1}\ y$
$du=\frac{1}{y\sqrt{y^2-1}}\ dy$
$dv=2y\ dy$
$v=y^2$
Dengan menggunakan integral parsial diperoleh
=$y^2sec^{-1}\ y-\int \frac{y}{\sqrt{y^2-1}}\ dy$
Misalkan : $z=y^2-1$
$dz=2y\ dy$
=$y^2sec^{-1}\ y-\int \frac{1}{2\sqrt{z}}\ dz$
=$y^2sec^{-1}\ y-\sqrt{z}+C$
=$y^2sec^{-1}\ y-\sqrt{y^2-1}+C$
=$xsec^{-1}\ \sqrt{x}-\sqrt{x-1}+C$
100) $\int x.\left(\frac{1-x^2}{1+x^2}\right)^{1/2}\ dx$
Jawab : $\int x.\left(\frac{1-x^2}{1+x^2}\right)^{1/2}\ dx$
Misalkan : $u=(1+x^2)^{1/2}$
$du=\frac{x}{(1+x^2)^{1/2}}\ dx$
=$\int (1-(u^2-1))^{1/2}\ du$
=$\int (2-u^2)^{1/2}\ du$
Substitusi : $u=\sqrt{2}sin\ y$
$du=\sqrt{2}cos\ y\ dy$
=$\int \sqrt{2(1-sin^2\ y)}.\sqrt{2}cos\ y\ dy$
=$\int 2cos^2\ y\ dy$
=$\int cos\ 2y+1\ dy$
=$\frac{1}{2}sin\ 2y+y+C$
=$sin\ y.cos\ y+y+C$
=$\frac{u}{\sqrt{2}}.\frac{\sqrt{2-u^2}}{\sqrt{2}}+arcsin\ \frac{u}{\sqrt{2}}+C$
=$\frac{\sqrt{1+x^2}}{\sqrt{2}}.\frac{\sqrt{1-x^2}}{\sqrt{2}}+arcsin\ \frac{\sqrt{1+x^2}}{\sqrt{2}}+C$
=$\frac{\sqrt{1-x^4}}{2}+arcsin\ \sqrt{\frac{1+x^2}{2}}+C$
Alhamdulillah pembahasannya selesai 😀😁
Saya pribadi meminta maaf jika ternyata masih banyak kekurangan dalam pembahasan ini. Apabila ada kritik dan masukan tuliskan di kolom komentar. Semoga pembahasannya bisa bermanfaat buat kalian semua...
Terima kasih telah berkunjung ke blog saya....🙏
Salam math lover";💕👌
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