Mathematics

# Solve :$\int{x}{e^x}\sin x dx$

##### SOLUTION
Now let $u=x$

$du=dx$

Now $\displaystyle dv=e^x\sin x\rightarrow v=\int e^x\sin x dx$

$\displaystyle \int e^x\sin x dx=\int \sin x d(e^x)$

Now $\displaystyle v=\int e^x\sin x=e^x\sin x-\int e^x d(\sin x)=e^x\sin x=\int e^x \cos x$

$\displaystyle=e^x\sin x-e^x\cos x-\int e^x\sin xdx$

$v=e^x\sin x-e^x\cos x-v$

$\Rightarrow 2v=e^x\sin x-e^x\cos x-v$

$\Rightarrow v=\dfrac{e^x}{2}(\sin x-\cos x)$

Now $\displaystyle \int xe^x\sin dx$ Let $u=x\Rightarrow du=dx$

and $v=\dfrac{e^x}{2}(\sin x-\cos x)$

$\displaystyle \int e^x\cos xdx=\dfrac{e^x}{2}(\sin x+\cos x)$

$\displaystyle\therefore \int xe^x\sin xdx=\dfrac{xe^x}{2}(\sin x-\cos x)-\int \dfrac{e^x}{2}(\sin x-\cos x)dx$

$\displaystyle =\dfrac{xe^x}{2}(\sin x-\cos x)-\dfrac{-1}{2}\int e^x\sin x+\dfrac{1}{2}\int e^x\cos dx$

$\displaystyle =\dfrac{xe^x}{2}(\sin x-\cos x)-\dfrac{e^x}{4}(\sin x-\cos x)+\dfrac{e^x}{4}(\sin x+\cos x)+c$

$\displaystyle I=\dfrac{xe^x}{2}(\sin x-\cos x)+\dfrac{e^x}{2}\cos x+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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