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
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