Question

Coba selesaikan integral parsial 3 suku ini
∫ xeˣ sin x dx

Answer 1

Verified answer

Jawab:

Penjelasan dengan langkah-langkah:

Integral

sifat integral parsial

[tex]\boxed{\displaystyle\int~uv'~dx\Rightarrow~uv-\int~vu'dx}[/tex]

Soal

[tex]\displaystyle\int~xe^x~\sin(x)~dx[/tex]

• u = x
• du = dx
• v' = [tex]\displaystyle e^x~\sin(x)[/tex]
• v =  [tex]\displaystyle~-\frac{1}{2}e^x\cos(x)+\frac{1}{2}e^x\sin(x)[/tex]

Maka

[tex]\begin{aligned}&=\displaystyle\int~xe^x~\sin(x)~dx\\&=x\left[\displaystyle~-\frac{1}{2}e^x\cos(x)+\frac{1}{2}e^x\sin(x)\right]-\int~\left[\displaystyle~-\frac{1}{2}e^x\cos(x)+\frac{1}{2}e^x\sin(x)\right]~dx\\&=\left[\displaystyle~-\frac{1}{2}xe^x\cos(x)+\frac{1}{2}xe^x\sin(x)\right]+\frac{1}{2}\int~e^x\cos(x)~dx-\frac{1}{2}\int~e^x\sin(x)~dx\end{aligned}[/tex]

• [tex]\int~e^x\cos(x)~dx\Rightarrow\displaystyle~\frac{1}{2}e^x\sin(x)+\frac{1}{2}e^x\cos(x)[/tex]

• [tex]\int~e^x\sin(x)~dx\Rightarrow-\frac{1}{2}e^x\cos(x)+\frac{1}{2}e^x\sin(x)[/tex]

[tex]\begin{aligned}&=\displaystyle~-\frac{1}{2}xe^x\cos(x)+\frac{1}{2}xe^x\sin(x)+\frac{1}{2}\left[\displaystyle~\frac{1}{2}e^x\sin(x)+\frac{1}{2}e^x\cos(x)\right]-\frac{1}{2}\left[-\frac{1}{2}e^x\cos(x)+\frac{1}{2}e^x\sin(x)\right]\\&=-\frac{1}{2}xe^x\cos(x)+\frac{1}{2}xe^x\sin(x)+\cancel{\frac{1}{4}e^x\sin(x)}+\frac{1}{4}e^x\cos(x)+\frac{1}{4}e^x\cos(x)\cancel{-\frac{1}{4}e^x\sin(x)}\\&=-\frac{1}{2}xe^x\cos(x)+\frac{1}{2}xe^x\sin(x)+\frac{2}{4}e^x\cos(x)\end{aligned}[/tex]

[tex]\begin{aligned}\displaystyle&=-\frac{1}{2}xe^x\cos(x)+\frac{1}{2}xe^x\sin(x)+\frac{2}{4}e^x\cos(x)\\&=\frac{1}{2}e^x\left[-x\cos(x)+x\sin(x)+\cos(x)\right]\\&=\boxed{\frac{1}{2}e^x\left[\cos(x)(1-x)+x\sin(x)\right]}\end{aligned}[/tex]