100 Integral Problems and Solutions (part 3)
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 51-75
Ini adalah pembahasan part ke 3, yaitu soal no 51-75
51) $\int\frac{1}{4+5cos\ \theta}\ d\theta$
Jawab : $\int\frac{1}{4+5cos\ \theta}\ d\theta$
=$\int\frac{1}{4+5\frac{(1-tan^2\ \frac{\theta}{2})}{sec^2\ \frac{\theta}{2}}}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{4(1+tan^2\ \frac{\theta}{2})+5(1-tan^2\ \frac{\theta}{2})}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{9-tan^2\ \frac{\theta}{2}}\ d\theta$
Misalkan : $u=tan\ \frac{\theta}{2}$
$du=\frac{sec^2\ \frac{\theta}{2}}{2}\ d\theta$
=$2\int\frac{1}{9-u^2}\ du$
=$2\int\frac{1}{(3-u)(3+u)}\ du$
=$2\int\frac{1}{6}\left(\frac{1}{3-u} +\frac{1}{3+u}\right)\ du$
=$\frac{1}{3}\left(-ln|u-3|+ln|u+3|\right)+C$
=$\frac{1}{3}\left(ln|tan\ \frac{\theta}{2}+3|-ln|tan\ \frac{\theta}{2}-3|\right)+C$
52) $\int\frac{(1+x^{2/3})^{3/2}}{x^{1/3}}\ dx$
Jawab : $\int\frac{(1+x^{2/3})^{3/2}}{x^{1/3}}\ dx$
misalkan : $u=1+x^{2/3}$
$du=\frac{2}{3x^{1/3}}\ dx$
=$\int{3u^{3/2}}{2}\ du$
=$\frac{3}{5}u^{5/2}+C$
=$\frac{3}{5}(1+x^{2/3})^{5/2}+C$
53) $\int\frac{(arcsin\ x)^2}{\sqrt{1-x^2}}\ dx$
Jawab : $\int\frac{(arcsin\ x)^2}{\sqrt{1-x^2}}\ dx$
misalkan : $x=sin\ u$
$dx=cos\ u\ du$
=$\int\frac{u^2}{\sqrt{1-sin^2\ u}}cos\ u\ du$
=$\int u^2\ du$
=$\frac{1}{3}u^3+C$
=$\frac{1}{3}arcsin^3\ x+C$
54) $\int\frac{1}{x^{3/2}(1+x^{1/3})}\ dx$
Jawab : Soal ekuivalen dengan
$\int\frac{1}{x^{5/6}}.\frac{1}{x^{2/3}(1+x^{1/3})}\ dx$
Substitusi $u=x^{1/6}$
$du=\frac{1}{6x^{5/6}}\ dx$
=$6\int\frac{1}{u^4(1+u^2)}\ du$
=$6\int\left(\frac{-u^2+1}{u^4}+\frac{1}{1+u^2}\right)\ du$
=$6\int(-u^{-2}+u^{-4}+\frac{1}{1+u^2})\ du$
=$6u^{-1}-2u^{-3}+arctan\ u+C$
=$\frac{6}{x^{1/6}}-\frac{2}{x^{1/2}}+arctan\ (x^{1/6})+C$
55) $\int tan^3\ z\ dz$
Jawab : $\int tan^3\ z\ dz$
=$\int tan\ z(sec^2\ z-1)\ dz$
=$\int tan\ z.sec^2\ z\ dz-\int tan\ z\ dz$
misalkan : $u=tan\ z$
$du=sec^2\ z\ dz$
=$\int u\ du-ln|sec\ z|$
=$\frac{u^2}{2}-ln|sec\ z|+C$
=$\frac{tan^2\ z}{2}-ln|sec\ z|+C$
56) $\int sin^2\ \omega\ cos^4\ \omega\ d\omega$
Jawab : $\int sin^2\ \omega\ cos^4\ \omega\ d\omega$
=$\int \left(\frac{sin^2\ 2\omega}{4}\right)\left(cos^2\ \omega\right)\ d\omega$
=$\int \left(\frac{1-cos\ 4\omega}{8}\right)\left(\frac{cos\ 2\omega+1}{2}\right)\ d\omega$
=$\int \frac{cos\ 2\omega +1-cos\ 4\omega .cos\ 2\omega-cos\ 4\omega}{16}\ d\omega$
=$\int \frac{cos\ 2\omega +1-\frac{cos\ 6\omega+cos\ 2\omega}{2}-cos\ 4\omega}{16}\ d\omega$
=$\int \frac{-\frac{cos\ 6\omega}{2}-cos\ 4\omega+\frac{cos\ 2\omega}{2}+1}{16}\ d\omega$
=$\frac{1}{16}\left(-\frac{sin\ 6\omega}{12}-\frac{sin\ 4\omega}{4}+\frac{sin\ 2\omega}{4}+\omega \right)+C$
=$-\frac{sin\ 6\omega}{192}-\frac{sin\ 4\omega}{64}+\frac{sin\ 2\omega}{64}+\omega+C$
57) $\int\frac{xe^{x^2}}{1+e^{2x^2}}\ dx$
Jawab : $\int\frac{xe^{x^2}}{1+e^{2x^2}}\ dx$
misalkan : $u=e^{x^2}$
$du=2xe^{x^2}\ dx$
=$\frac{1}{2}\int\frac{1}{1+u^2}\ du$
=$\frac{1}{2}arctan\ u+C$
=$\frac{1}{2}arctan\ (e^{x^2})+C$
58) $\int\frac{cos^3\ x}{\sqrt{sin\ x}}\ dx$
Jawab : $\int\frac{cos^3\ x}{\sqrt{sin\ x}}\ dx$
=$\int\frac{(1-sin^2\ x)cos\ x}{\sqrt{sin\ x}}\ dx$
misalkan $u=sin\ x$
$du=cos\ x\ dx$
$\int\frac{1-u^2}{\sqrt{u}}\ du$
=$\int\frac{1}{\sqrt{u}}-u\sqrt{u}\ du$
=$2\sqrt{u}-\frac{2}{5}u^2\sqrt{u}+C$
=$2\sqrt{sin\ x}-\frac{2}{5}sin^2\ x.\sqrt{sin\ x}+C$
59) $\int x^3exp(-x^2)\ dx$
Jawab : $\int x^3e^{-x^2}\ dx$
misalkan : $u=e^{-x^2}$
$du=-2xe^{-x^2}\ dx$
=$\int \frac{-ln|u|}{-2}\ du$
=$\frac{1}{2}\int ln|u|\ du$
Dengan menggunakan integral parsial diperoleh
=$\frac{1}{2}(u.ln|u|-u)+C$
=$\frac{e^{-x^2}.(-x^2)-e^{-x^2}}{2}+C$
60) $\int sin\ \sqrt{x}\ dx$
Jawab : $\int sin\ \sqrt{x}\ dx$
Substitusi : $x=u^2$
$dx=2u\ du$
=$\int sin\ u.2u\ du$
Dengan menggunakan integral parsial
=$2u.(-cos\ u)-\int (-cos\ u)2\ du$
=$-2u.cos\ u+2sin\ u+C$
=$-2\sqrt{x}.cos\ \sqrt{x}+2sin\ \sqrt{x}+C$
61) $\int\frac{arcsin\ x}{x^2}\ dx$
Jawab : $\int\frac{arcsin\ x}{x^2}\ dx$
Misalkan : $u=arcsin\ x$
$du=\frac{1}{\sqrt{1-x^2}}\ dx$
$dv=\frac{1}{x^2}\ dx$
$v=-\frac{1}{x}$
Dengan menggunakan integral parsial
=$uv-\int v\ du$
=$arcsin\ x(-\frac{1}{x})-\int (-\frac{1}{x(1-x^2)})\ dx$
=$-\frac{arcsin\ x}{x}-\int (-\frac{1}{x(1-x)(1+x)})\ dx$
=$-\frac{arcsin\ x}{x}-\int (\frac{1}{x(x-1)(x+1)})\ dx$
=$-\frac{arcsin\ x}{x}-\int (\frac{1}{2(x-1)}-\frac{1}{x}+\frac{1}{2(x+1)})\ dx$
=$-\frac{arcsin\ x}{x}-\frac{1}{2}ln|x-1|$ $+ln|x|-\frac{1}{2}ln|x+1|+C$
62) $\int\sqrt{x^2-9}\ dx$
Jawab : $\int\sqrt{x^2-9}\ dx$
Substitusi : $x=3sec\ u$
$dx=3sec\ u.tan\ u\ du$
=$\int\sqrt{9sec^2\ u-9}3sec\ u.tan\ u\ du$
=$9\int sec\ u.tan^2\ u\ du$
=$9\int sec^3\ u-sec\ u\ du$
Kita menggunakan formula
$\int sec^n\ x\ dx=sec^{n-2}\ x.tan\ x$ $-(n-2)\int sec^n\ x\ dx$ $+(n-2)\int sec^{n-2}\ x\ dx$
maka dengan formula tsb nanti akan diperoleh
$\int sec^3\ x=\frac{1}{2}.sec\ x.tan\ x$ $+\frac{1}{2}ln|sec\ x+tan\ x|+C$
dan kita tau
$\int sec\ x=ln|sec\ x+tan\ x|+C$
Jadi $9\int sec^3\ u-sec\ u\ du$
=$\frac{9sec\ x.tan\ x}{2}-\frac{9ln|sec\ x+tan\ x|}{2}+C$
63) $\int x^2\sqrt{1-x^2}\ dx$
Jawab : $\int x^2\sqrt{1-x^2}\ dx$
Substitusi : $x=sin\ y$
$dx=cos\ y\ dy$
=$\int sin^2\ y\sqrt{1-sin^2\ y}cos\ y\ dy$
=$\int sin^2\ y.cos^2\ y\ dy$
=$\int\frac{sin^2\ 2y}{4}\ dy$
=$\int\frac{1-cos\ 4y}{8}\ dy$
=$\frac{y}{8}-\frac{sin\ 4y}{32}+C$
=$\frac{y}{8}-\frac{sin\ 2y.cos\ 2y}{16}+C$
=$\frac{y}{8}-\frac{sin\ y.cos\ y.(1-2sin^2\ y)}{8}+C$
=$\frac{y}{8}-\frac{sin\ y.cos\ y.(1-2sin^2\ y)}{8}+C$
64) $\int x\sqrt{2x-x^2}\ dx$
Jawab : $\int x\sqrt{2x-x^2}\ dx$
=$\int x\sqrt{-(x-1)^2+1}\ dx$
Substitusi : $x-1=sin\ u$
$dx=cos\ u\ du$
=$\int (sin\ u+1)cos^2\ u\ du$
=$\int (sin\ u+1)\frac{(cos\ 2u+1)}{2}\ du$
=$\int \frac{sin\ u.cos\ 2u+sin\ u+cos\ 2u+1}{2}\ du$
=$\frac{1}{2}\int \frac{sin\ 3u-sin\ u}{2}+sin\ u+cos\ 2u+1\ du$
=$\frac{1}{2}\int \frac{sin\ 3u+sin\ u}{2}+cos\ 2u+1\ du$
=$-\frac{1}{12}cos\ 3u-\frac{1}{4}cos\ u+\frac{1}{2}sin\ 2u+u+C$
=$-\frac{cos\ u.cos\ 2u-sin\ u.sin\ 2u}{12}$ $-\frac{1}{4}cos\ u+sin\ u.cos\ u+u+C$
=$-\frac{cos\ u.(1-2sin^2\ u)-2sin^2\ u.cos\ u}{12}$ $-\frac{1}{4}cos\ u+sin\ u.cos\ u+u+C$
=$-\frac{cos\ u.(-2sin^2\ u)-2sin^2\ u.cos\ u}{12}$ $-\frac{1}{3}cos\ u+sin\ u.cos\ u+u+C$
=$\frac{sin^2\ u.cos\ u}{3}-\frac{cos\ u}{3}+sin\ u.cos\ u+u+C$
=$\frac{(x-1)^2.\sqrt{2x-x^2}}{3}-\frac{\sqrt{2x-x^2}}{3}+$ $(x-1)\sqrt{2x-x^2}+arcsin\ (x-1)+C$
65) $\int\frac{x-2}{4x^2+4x+1}\ dx$
Jawab : $\int\frac{x-2}{4x^2+4x+1}\ dx$
=$\int\frac{x-2}{(2x+1)^2}\ dx$
=$\frac{1}{2}\int\frac{2x+1-5}{(2x+1)^2}\ dx$
=$\frac{1}{2}\int\frac{1}{2x+1}-\frac{5}{(2x+1)^2}\ dx$
=$\frac{1}{4}ln|2x+1|+\frac{5}{2(2x+1)}+C$
=$\frac{1}{4}ln|2x+1|+\frac{5}{4x+2}+C$
66) $\int\frac{2x^2-5x-1}{x^3-2x^2-x+2}\ dx$
Jawab : $\int\frac{2x^2-5x-1}{x^3-2x^2-x+2}\ dx$
=$\int\frac{2x^2-5x-1}{(x-1)(x-2)(x+1)}\ dx$
=$\int\frac{2}{x-1}-\frac{1}{x-2}+\frac{1}{x+1}\ dx$
=$2ln|x-1|-ln|x-2|+ln|x+1|+C$
67) $\int\frac{e^{2x}}{e^{2x}-1}\ dx$
Jawab : $\int\frac{e^{2x}}{e^{2x}-1}\ dx$
misalkan : $u=e^{2x}-1$
$du=2e^{2x}\ dx$
=$\int\frac{1}{2u}\ du$
=$\frac{1}{2}ln|u|+C$
=$\frac{1}{2}ln|e^{2x}-1|+C$
68) $\int\frac{cos\ x}{sin^2\ x-3sin\ x+2}\ dx$
Jawab : $\int\frac{cos\ x}{sin^2\ x-3sin\ x+2}\ dx$
=$\int\frac{cos\ x}{(sin\ x-2)(sin\ x -1)}\ dx$
misalkan : $u=sin\ x-2$
$du=cos\ x\ dx$
=$\int\frac{1}{u(u+1)}\ du$
=$\int\frac{1}{u}-\frac{1}{u+1}\ du$
=$ln|u|-ln|u+1|+C$
=$ln\left|\frac{u}{u+1}\right|+C$
=$ln\left|\frac{sin\ x-2}{sin\ x-1}\right|+C$
69) $\int\frac{2x^3+3x^2+4}{(x+1)^4}\ dx$
Jawab : $\int\frac{2x^3+3x^2+4}{(x+1)^4}\ dx$
=$\int\frac{2}{x+1}-\frac{3}{(x+1)^2}+\frac{5}{(x+1)^4}\ dx$
=$2ln|x+1|+\frac{3}{x+1}-\frac{5}{3(x+1)^3}+C$
70) $\int\frac{sec^2\ x}{tan^2\ x+2tan\ x+2}\ dx$
Jawab : $\int\frac{sec^2\ x}{tan^2\ x+2tan\ x+2}\ dx$
Substitusi : $u=tan\ x$
$du=sec^2\ x\ dx$
=$\int\frac{1}{u^2+2u+2}\ du$
=$\int\frac{1}{(u+1)^2+1}\ du$
=$arctan\ (u+1)+C$
=$arctan\ (tan\ x+1)+C$
71) $\int\frac{x^3+x^2+2x+1}{x^4+2x^2+1}\ dx$
Jawab : $\int\frac{x^3+x^2+2x+1}{x^4+2x^2+1}\ dx$
=$\int\frac{x^3+x^2+2x+1}{(x^2+1)^2}\ dx$
=$\int\frac{x(x^2+1)+(x^2+1)+x}{(x^2+1)^2}\ dx$
=$\int\frac{x}{x^2+1}+\frac{1}{x^2+1}+\frac{x}{(x^2+1)^2}\ dx$
=$\frac{1}{2}ln|x^2+1|+arctan\ x-\frac{1}{2(x^2+1)}+C$
=$\frac{1}{2}ln|x^2+1|+arctan\ x-\frac{1}{2x^2+2}+C$
72) $\int\frac{3+cos\ \theta}{2-cos\ \theta}\ d\theta$
Jawab : $\int\frac{3+cos\ \theta}{2-cos\ \theta}\ d\theta$
=$\int -1+\frac{5}{2-cos\ \theta}\ d\theta$
=$-\theta+5\int \frac{1}{1+2sin^2\ \frac{\theta}{2}}\ d\theta$
=$-\theta+5\int\frac{sec^2\ \frac{\theta}{2}}{sec^2\ \frac{\theta}{2}+2tan^2\ \frac{\theta}{2}}\ d\theta$
=$-\theta+5\int\frac{sec^2\ \frac{\theta}{2}}{1+3tan^2\ \frac{\theta}{2}}\ d\theta$
misalkan : $u=\sqrt{3}tan\ \frac{\theta}{2}$
$du=\sqrt{3}sec^2\ \frac{\theta}{2}\ d\theta$
=$-\theta+5\int\frac{1}{\sqrt{3}(1+u^2)}\ du$
=$-\theta+\frac{5\sqrt{3}}{3}arctan\ u +C$
=$-\theta+\frac{5\sqrt{3}}{3}arctan\ \left(\sqrt{3}tan\ \frac{\theta}{2}\right)+C$
73) $\int x^5\sqrt{x^3-1}\ dx$
Jawab : $\int x^5\sqrt{x^3-1}\ dx$
misalkan :$u=x^3-1$
$du=3x^2\ dx$
=$\int\frac{(u+1)\sqrt{u}}{3}\ du$
=$\frac{2}{15}u^{\frac{5}{2}}+\frac{2}{9}u^{\frac{3}{2}}+C$
=$\frac{2}{15}(x^3-1)^{\frac{5}{2}}+\frac{2}{9}(x^3-1)^{\frac{3}{2}}+C$
74) $\int\frac{1}{2+2cos\ \theta+sin\ \theta}\ d\theta$
Jawab : $\int\frac{1}{2+2cos\ \theta+sin\ \theta}\ d\theta$
=$\int\frac{1}{2+4cos^2\ \frac{\theta}{2}-2+2sin\ \frac{\theta}{2}.cos\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{1}{2+4cos^2\ \frac{\theta}{2}-2+2sin\ \frac{\theta}{2}.cos\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{1}{4cos^2\ \frac{\theta}{2}+2sin\ \frac{\theta}{2}.cos\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{4+2tan\ \frac{\theta}{2}}\ d\theta$
Misalkan : $u=tan\ \frac{\theta}{2}$
$du=\frac{sec^2\ \frac{\theta}{2}}{2}\ d\theta$
=$\int\frac{2}{4+2u}\ du$
=$ln|u+2|+C$
=$ln|tan\ \frac{\theta}{2}+2|+C$
75) $\int\frac{\sqrt{1+sin\ x}}{sec\ x}\ dx$
Jawab : $\int\frac{\sqrt{1+sin\ x}}{sec\ x}\ dx$
=$\int cos\ x.\sqrt{1+sin\ x}\ dx$
misalkan : $u=1+sin\ x$
$du=cos\ x\ dx$
=$\int\sqrt{u}\ du$
=$\frac{2}{3}u\sqrt{u}+C$
=$\frac{2}{3}(1+sin\ x)\sqrt{1+sin\ x}+C$
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";💕👌
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 51-75
 |
| Soal Integral Aljabar dan Trigonometri |
Ini adalah pembahasan part ke 3, yaitu soal no 51-75
51) $\int\frac{1}{4+5cos\ \theta}\ d\theta$
Jawab : $\int\frac{1}{4+5cos\ \theta}\ d\theta$
=$\int\frac{1}{4+5\frac{(1-tan^2\ \frac{\theta}{2})}{sec^2\ \frac{\theta}{2}}}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{4(1+tan^2\ \frac{\theta}{2})+5(1-tan^2\ \frac{\theta}{2})}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{9-tan^2\ \frac{\theta}{2}}\ d\theta$
Misalkan : $u=tan\ \frac{\theta}{2}$
$du=\frac{sec^2\ \frac{\theta}{2}}{2}\ d\theta$
=$2\int\frac{1}{9-u^2}\ du$
=$2\int\frac{1}{(3-u)(3+u)}\ du$
=$2\int\frac{1}{6}\left(\frac{1}{3-u} +\frac{1}{3+u}\right)\ du$
=$\frac{1}{3}\left(-ln|u-3|+ln|u+3|\right)+C$
=$\frac{1}{3}\left(ln|tan\ \frac{\theta}{2}+3|-ln|tan\ \frac{\theta}{2}-3|\right)+C$
52) $\int\frac{(1+x^{2/3})^{3/2}}{x^{1/3}}\ dx$
Jawab : $\int\frac{(1+x^{2/3})^{3/2}}{x^{1/3}}\ dx$
misalkan : $u=1+x^{2/3}$
$du=\frac{2}{3x^{1/3}}\ dx$
=$\int{3u^{3/2}}{2}\ du$
=$\frac{3}{5}u^{5/2}+C$
=$\frac{3}{5}(1+x^{2/3})^{5/2}+C$
53) $\int\frac{(arcsin\ x)^2}{\sqrt{1-x^2}}\ dx$
Jawab : $\int\frac{(arcsin\ x)^2}{\sqrt{1-x^2}}\ dx$
misalkan : $x=sin\ u$
$dx=cos\ u\ du$
=$\int\frac{u^2}{\sqrt{1-sin^2\ u}}cos\ u\ du$
=$\int u^2\ du$
=$\frac{1}{3}u^3+C$
=$\frac{1}{3}arcsin^3\ x+C$
54) $\int\frac{1}{x^{3/2}(1+x^{1/3})}\ dx$
Jawab : Soal ekuivalen dengan
$\int\frac{1}{x^{5/6}}.\frac{1}{x^{2/3}(1+x^{1/3})}\ dx$
Substitusi $u=x^{1/6}$
$du=\frac{1}{6x^{5/6}}\ dx$
=$6\int\frac{1}{u^4(1+u^2)}\ du$
=$6\int\left(\frac{-u^2+1}{u^4}+\frac{1}{1+u^2}\right)\ du$
=$6\int(-u^{-2}+u^{-4}+\frac{1}{1+u^2})\ du$
=$6u^{-1}-2u^{-3}+arctan\ u+C$
=$\frac{6}{x^{1/6}}-\frac{2}{x^{1/2}}+arctan\ (x^{1/6})+C$
55) $\int tan^3\ z\ dz$
Jawab : $\int tan^3\ z\ dz$
=$\int tan\ z(sec^2\ z-1)\ dz$
=$\int tan\ z.sec^2\ z\ dz-\int tan\ z\ dz$
misalkan : $u=tan\ z$
$du=sec^2\ z\ dz$
=$\int u\ du-ln|sec\ z|$
=$\frac{u^2}{2}-ln|sec\ z|+C$
=$\frac{tan^2\ z}{2}-ln|sec\ z|+C$
56) $\int sin^2\ \omega\ cos^4\ \omega\ d\omega$
Jawab : $\int sin^2\ \omega\ cos^4\ \omega\ d\omega$
=$\int \left(\frac{sin^2\ 2\omega}{4}\right)\left(cos^2\ \omega\right)\ d\omega$
=$\int \left(\frac{1-cos\ 4\omega}{8}\right)\left(\frac{cos\ 2\omega+1}{2}\right)\ d\omega$
=$\int \frac{cos\ 2\omega +1-cos\ 4\omega .cos\ 2\omega-cos\ 4\omega}{16}\ d\omega$
=$\int \frac{cos\ 2\omega +1-\frac{cos\ 6\omega+cos\ 2\omega}{2}-cos\ 4\omega}{16}\ d\omega$
=$\int \frac{-\frac{cos\ 6\omega}{2}-cos\ 4\omega+\frac{cos\ 2\omega}{2}+1}{16}\ d\omega$
=$\frac{1}{16}\left(-\frac{sin\ 6\omega}{12}-\frac{sin\ 4\omega}{4}+\frac{sin\ 2\omega}{4}+\omega \right)+C$
=$-\frac{sin\ 6\omega}{192}-\frac{sin\ 4\omega}{64}+\frac{sin\ 2\omega}{64}+\omega+C$
57) $\int\frac{xe^{x^2}}{1+e^{2x^2}}\ dx$
Jawab : $\int\frac{xe^{x^2}}{1+e^{2x^2}}\ dx$
misalkan : $u=e^{x^2}$
$du=2xe^{x^2}\ dx$
=$\frac{1}{2}\int\frac{1}{1+u^2}\ du$
=$\frac{1}{2}arctan\ u+C$
=$\frac{1}{2}arctan\ (e^{x^2})+C$
58) $\int\frac{cos^3\ x}{\sqrt{sin\ x}}\ dx$
Jawab : $\int\frac{cos^3\ x}{\sqrt{sin\ x}}\ dx$
=$\int\frac{(1-sin^2\ x)cos\ x}{\sqrt{sin\ x}}\ dx$
misalkan $u=sin\ x$
$du=cos\ x\ dx$
$\int\frac{1-u^2}{\sqrt{u}}\ du$
=$\int\frac{1}{\sqrt{u}}-u\sqrt{u}\ du$
=$2\sqrt{u}-\frac{2}{5}u^2\sqrt{u}+C$
=$2\sqrt{sin\ x}-\frac{2}{5}sin^2\ x.\sqrt{sin\ x}+C$
59) $\int x^3exp(-x^2)\ dx$
Jawab : $\int x^3e^{-x^2}\ dx$
misalkan : $u=e^{-x^2}$
$du=-2xe^{-x^2}\ dx$
=$\int \frac{-ln|u|}{-2}\ du$
=$\frac{1}{2}\int ln|u|\ du$
Dengan menggunakan integral parsial diperoleh
=$\frac{1}{2}(u.ln|u|-u)+C$
=$\frac{e^{-x^2}.(-x^2)-e^{-x^2}}{2}+C$
60) $\int sin\ \sqrt{x}\ dx$
Jawab : $\int sin\ \sqrt{x}\ dx$
Substitusi : $x=u^2$
$dx=2u\ du$
=$\int sin\ u.2u\ du$
Dengan menggunakan integral parsial
=$2u.(-cos\ u)-\int (-cos\ u)2\ du$
=$-2u.cos\ u+2sin\ u+C$
=$-2\sqrt{x}.cos\ \sqrt{x}+2sin\ \sqrt{x}+C$
61) $\int\frac{arcsin\ x}{x^2}\ dx$
Jawab : $\int\frac{arcsin\ x}{x^2}\ dx$
Misalkan : $u=arcsin\ x$
$du=\frac{1}{\sqrt{1-x^2}}\ dx$
$dv=\frac{1}{x^2}\ dx$
$v=-\frac{1}{x}$
Dengan menggunakan integral parsial
=$uv-\int v\ du$
=$arcsin\ x(-\frac{1}{x})-\int (-\frac{1}{x(1-x^2)})\ dx$
=$-\frac{arcsin\ x}{x}-\int (-\frac{1}{x(1-x)(1+x)})\ dx$
=$-\frac{arcsin\ x}{x}-\int (\frac{1}{x(x-1)(x+1)})\ dx$
=$-\frac{arcsin\ x}{x}-\int (\frac{1}{2(x-1)}-\frac{1}{x}+\frac{1}{2(x+1)})\ dx$
=$-\frac{arcsin\ x}{x}-\frac{1}{2}ln|x-1|$ $+ln|x|-\frac{1}{2}ln|x+1|+C$
62) $\int\sqrt{x^2-9}\ dx$
Jawab : $\int\sqrt{x^2-9}\ dx$
Substitusi : $x=3sec\ u$
$dx=3sec\ u.tan\ u\ du$
=$\int\sqrt{9sec^2\ u-9}3sec\ u.tan\ u\ du$
=$9\int sec\ u.tan^2\ u\ du$
=$9\int sec^3\ u-sec\ u\ du$
Kita menggunakan formula
$\int sec^n\ x\ dx=sec^{n-2}\ x.tan\ x$ $-(n-2)\int sec^n\ x\ dx$ $+(n-2)\int sec^{n-2}\ x\ dx$
maka dengan formula tsb nanti akan diperoleh
$\int sec^3\ x=\frac{1}{2}.sec\ x.tan\ x$ $+\frac{1}{2}ln|sec\ x+tan\ x|+C$
dan kita tau
$\int sec\ x=ln|sec\ x+tan\ x|+C$
Jadi $9\int sec^3\ u-sec\ u\ du$
=$\frac{9sec\ x.tan\ x}{2}-\frac{9ln|sec\ x+tan\ x|}{2}+C$
63) $\int x^2\sqrt{1-x^2}\ dx$
Jawab : $\int x^2\sqrt{1-x^2}\ dx$
Substitusi : $x=sin\ y$
$dx=cos\ y\ dy$
=$\int sin^2\ y\sqrt{1-sin^2\ y}cos\ y\ dy$
=$\int sin^2\ y.cos^2\ y\ dy$
=$\int\frac{sin^2\ 2y}{4}\ dy$
=$\int\frac{1-cos\ 4y}{8}\ dy$
=$\frac{y}{8}-\frac{sin\ 4y}{32}+C$
=$\frac{y}{8}-\frac{sin\ 2y.cos\ 2y}{16}+C$
=$\frac{y}{8}-\frac{sin\ y.cos\ y.(1-2sin^2\ y)}{8}+C$
=$\frac{y}{8}-\frac{sin\ y.cos\ y.(1-2sin^2\ y)}{8}+C$
64) $\int x\sqrt{2x-x^2}\ dx$
Jawab : $\int x\sqrt{2x-x^2}\ dx$
=$\int x\sqrt{-(x-1)^2+1}\ dx$
Substitusi : $x-1=sin\ u$
$dx=cos\ u\ du$
=$\int (sin\ u+1)cos^2\ u\ du$
=$\int (sin\ u+1)\frac{(cos\ 2u+1)}{2}\ du$
=$\int \frac{sin\ u.cos\ 2u+sin\ u+cos\ 2u+1}{2}\ du$
=$\frac{1}{2}\int \frac{sin\ 3u-sin\ u}{2}+sin\ u+cos\ 2u+1\ du$
=$\frac{1}{2}\int \frac{sin\ 3u+sin\ u}{2}+cos\ 2u+1\ du$
=$-\frac{1}{12}cos\ 3u-\frac{1}{4}cos\ u+\frac{1}{2}sin\ 2u+u+C$
=$-\frac{cos\ u.cos\ 2u-sin\ u.sin\ 2u}{12}$ $-\frac{1}{4}cos\ u+sin\ u.cos\ u+u+C$
=$-\frac{cos\ u.(1-2sin^2\ u)-2sin^2\ u.cos\ u}{12}$ $-\frac{1}{4}cos\ u+sin\ u.cos\ u+u+C$
=$-\frac{cos\ u.(-2sin^2\ u)-2sin^2\ u.cos\ u}{12}$ $-\frac{1}{3}cos\ u+sin\ u.cos\ u+u+C$
=$\frac{sin^2\ u.cos\ u}{3}-\frac{cos\ u}{3}+sin\ u.cos\ u+u+C$
=$\frac{(x-1)^2.\sqrt{2x-x^2}}{3}-\frac{\sqrt{2x-x^2}}{3}+$ $(x-1)\sqrt{2x-x^2}+arcsin\ (x-1)+C$
65) $\int\frac{x-2}{4x^2+4x+1}\ dx$
Jawab : $\int\frac{x-2}{4x^2+4x+1}\ dx$
=$\int\frac{x-2}{(2x+1)^2}\ dx$
=$\frac{1}{2}\int\frac{2x+1-5}{(2x+1)^2}\ dx$
=$\frac{1}{2}\int\frac{1}{2x+1}-\frac{5}{(2x+1)^2}\ dx$
=$\frac{1}{4}ln|2x+1|+\frac{5}{2(2x+1)}+C$
=$\frac{1}{4}ln|2x+1|+\frac{5}{4x+2}+C$
66) $\int\frac{2x^2-5x-1}{x^3-2x^2-x+2}\ dx$
Jawab : $\int\frac{2x^2-5x-1}{x^3-2x^2-x+2}\ dx$
=$\int\frac{2x^2-5x-1}{(x-1)(x-2)(x+1)}\ dx$
=$\int\frac{2}{x-1}-\frac{1}{x-2}+\frac{1}{x+1}\ dx$
=$2ln|x-1|-ln|x-2|+ln|x+1|+C$
67) $\int\frac{e^{2x}}{e^{2x}-1}\ dx$
Jawab : $\int\frac{e^{2x}}{e^{2x}-1}\ dx$
misalkan : $u=e^{2x}-1$
$du=2e^{2x}\ dx$
=$\int\frac{1}{2u}\ du$
=$\frac{1}{2}ln|u|+C$
=$\frac{1}{2}ln|e^{2x}-1|+C$
68) $\int\frac{cos\ x}{sin^2\ x-3sin\ x+2}\ dx$
Jawab : $\int\frac{cos\ x}{sin^2\ x-3sin\ x+2}\ dx$
=$\int\frac{cos\ x}{(sin\ x-2)(sin\ x -1)}\ dx$
misalkan : $u=sin\ x-2$
$du=cos\ x\ dx$
=$\int\frac{1}{u(u+1)}\ du$
=$\int\frac{1}{u}-\frac{1}{u+1}\ du$
=$ln|u|-ln|u+1|+C$
=$ln\left|\frac{u}{u+1}\right|+C$
=$ln\left|\frac{sin\ x-2}{sin\ x-1}\right|+C$
69) $\int\frac{2x^3+3x^2+4}{(x+1)^4}\ dx$
Jawab : $\int\frac{2x^3+3x^2+4}{(x+1)^4}\ dx$
=$\int\frac{2}{x+1}-\frac{3}{(x+1)^2}+\frac{5}{(x+1)^4}\ dx$
=$2ln|x+1|+\frac{3}{x+1}-\frac{5}{3(x+1)^3}+C$
70) $\int\frac{sec^2\ x}{tan^2\ x+2tan\ x+2}\ dx$
Jawab : $\int\frac{sec^2\ x}{tan^2\ x+2tan\ x+2}\ dx$
Substitusi : $u=tan\ x$
$du=sec^2\ x\ dx$
=$\int\frac{1}{u^2+2u+2}\ du$
=$\int\frac{1}{(u+1)^2+1}\ du$
=$arctan\ (u+1)+C$
=$arctan\ (tan\ x+1)+C$
71) $\int\frac{x^3+x^2+2x+1}{x^4+2x^2+1}\ dx$
Jawab : $\int\frac{x^3+x^2+2x+1}{x^4+2x^2+1}\ dx$
=$\int\frac{x^3+x^2+2x+1}{(x^2+1)^2}\ dx$
=$\int\frac{x(x^2+1)+(x^2+1)+x}{(x^2+1)^2}\ dx$
=$\int\frac{x}{x^2+1}+\frac{1}{x^2+1}+\frac{x}{(x^2+1)^2}\ dx$
=$\frac{1}{2}ln|x^2+1|+arctan\ x-\frac{1}{2(x^2+1)}+C$
=$\frac{1}{2}ln|x^2+1|+arctan\ x-\frac{1}{2x^2+2}+C$
72) $\int\frac{3+cos\ \theta}{2-cos\ \theta}\ d\theta$
Jawab : $\int\frac{3+cos\ \theta}{2-cos\ \theta}\ d\theta$
=$\int -1+\frac{5}{2-cos\ \theta}\ d\theta$
=$-\theta+5\int \frac{1}{1+2sin^2\ \frac{\theta}{2}}\ d\theta$
=$-\theta+5\int\frac{sec^2\ \frac{\theta}{2}}{sec^2\ \frac{\theta}{2}+2tan^2\ \frac{\theta}{2}}\ d\theta$
=$-\theta+5\int\frac{sec^2\ \frac{\theta}{2}}{1+3tan^2\ \frac{\theta}{2}}\ d\theta$
misalkan : $u=\sqrt{3}tan\ \frac{\theta}{2}$
$du=\sqrt{3}sec^2\ \frac{\theta}{2}\ d\theta$
=$-\theta+5\int\frac{1}{\sqrt{3}(1+u^2)}\ du$
=$-\theta+\frac{5\sqrt{3}}{3}arctan\ u +C$
=$-\theta+\frac{5\sqrt{3}}{3}arctan\ \left(\sqrt{3}tan\ \frac{\theta}{2}\right)+C$
73) $\int x^5\sqrt{x^3-1}\ dx$
Jawab : $\int x^5\sqrt{x^3-1}\ dx$
misalkan :$u=x^3-1$
$du=3x^2\ dx$
=$\int\frac{(u+1)\sqrt{u}}{3}\ du$
=$\frac{2}{15}u^{\frac{5}{2}}+\frac{2}{9}u^{\frac{3}{2}}+C$
=$\frac{2}{15}(x^3-1)^{\frac{5}{2}}+\frac{2}{9}(x^3-1)^{\frac{3}{2}}+C$
74) $\int\frac{1}{2+2cos\ \theta+sin\ \theta}\ d\theta$
Jawab : $\int\frac{1}{2+2cos\ \theta+sin\ \theta}\ d\theta$
=$\int\frac{1}{2+4cos^2\ \frac{\theta}{2}-2+2sin\ \frac{\theta}{2}.cos\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{1}{2+4cos^2\ \frac{\theta}{2}-2+2sin\ \frac{\theta}{2}.cos\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{1}{4cos^2\ \frac{\theta}{2}+2sin\ \frac{\theta}{2}.cos\ \frac{\theta}{2}}\ d\theta$
=$\int\frac{sec^2\ \frac{\theta}{2}}{4+2tan\ \frac{\theta}{2}}\ d\theta$
Misalkan : $u=tan\ \frac{\theta}{2}$
$du=\frac{sec^2\ \frac{\theta}{2}}{2}\ d\theta$
=$\int\frac{2}{4+2u}\ du$
=$ln|u+2|+C$
=$ln|tan\ \frac{\theta}{2}+2|+C$
75) $\int\frac{\sqrt{1+sin\ x}}{sec\ x}\ dx$
Jawab : $\int\frac{\sqrt{1+sin\ x}}{sec\ x}\ dx$
=$\int cos\ x.\sqrt{1+sin\ x}\ dx$
misalkan : $u=1+sin\ x$
$du=cos\ x\ dx$
=$\int\sqrt{u}\ du$
=$\frac{2}{3}u\sqrt{u}+C$
=$\frac{2}{3}(1+sin\ x)\sqrt{1+sin\ x}+C$
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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