1. ## Confusing integral

How do I start this?

2. Let $t=\sin^{-1}x\Rightarrow dt = \frac{dx}{\sqrt{1-x^2}}$ so $\int \frac{xe^{\sin^{-1}x}}{\sqrt{1-x^2}}dx=\int \sin x e^x dx=I$ and $I=e^x\sin x-\int \cos x e^xdx = e^x \sin x - (\cos xe^x + \int \sin x e^x dx)$ $=e^x(\sin x - \cos x)-I\Rightarrow I =e^x\frac{\sin x -\cos x }{2}+C$.

3. ^ I lost you at I ?

4. $I=e^x\sin x -e^x \cos x -I \Rightarrow 2I=e^x\sin x - e^x \cos x$

5. Hello, VkL!

Abu-Khalil is correct . . . (he could have been clearer)

$20)\;\;\int\frac{xe^{\sin^{-1}\!x}}{\sqrt{1-x^2}}\,dx$

We have: . $\int x\cdot e^{\sin^{1}x}\cdot\frac{dx}{\sqrt{1-x^2}}$

Let: . $t \:=\: \sin^{-1}\!x \quad\Rightarrow\quad dt \:=\:\frac{dx}{\sqrt{1-x^2}}$
Note that: . $x \:=\:\sin t$

Substitute: . $\int \sin t\cdot e^t\cdot dt$

We will integrate by parts: . $I \;=\;\int e^t\sin t\,dt$

. . $\begin{array}{ccccccc}u &=& \sin t & & dv &=& e^t\,dt \\ du &=& \cos t\,dt & & v &=& e^t\end{array}$

We have: . $I \;=\;e^t\sin t - \int e^t\cos t\,dt$

. . By parts: . $\begin{array}{ccccccc}u &=& \cos t & & dv&=& e^t\,dt \\ du &=& -\sin t\,dt & & v &=& e^t \end{array}$

And we have: . $I \;=\;e^t\sin t - \bigg[e^t\cos t + \int e^t\sin t\,dt\bigg]$

. . . . . . . . . $I \;=\;e^t\sin t - e^t\cos t - \underbrace{\int e^t\sin t\,dt}_{\text{This is }I} + C$

So we have: . $I \;=\;e^t\sin t - e^t\cos t - I + C$

. . . . . . . . . $2I \;=\;e^t\sin t - e^t\cos t + C$

. . . . . . . . . . $I \;=\;\tfrac{1}{2}\left(e^t\sin t - e^t\cos t\right) + C$

Hence: . $\int e^t\sin t\,dt \;=\;\tfrac{1}{2}\,e^t(\sin t - \cos t) + C$

Back-substitute: . $t \,=\,\sin^{-1}\!x$

We have: . $\frac{1}{2}e^{\sin^{-1}\!x}\bigg[\sin\left(\sin^{-1}\!{x}\right) - \cos\left(\sin^{-1}\!x\right)\bigg] + C$

. . . . . . . $= \;\frac{1}{2}\,e^{\sin^{-1}\!x}\bigg[x - \sqrt{1-x^2}\bigg] + C$