Mathematics

Evaluate:
$$\displaystyle \int e^{2{\sin^{-1}x}}dx$$


SOLUTION
$$I=\displaystyle\int e^{2\sin^{-1}x}dx$$

Let $$u=\sin^{-1}x\Rightarrow \sin u =x \Rightarrow\cos u= \sqrt{1-x^2} $$

$$du=\dfrac{1}{\sqrt{1-x^{2}}}dx$$

$$dx=du\sqrt{1-x^{2}}=\cos u\ du$$

$$I=\displaystyle\int e^{2u}\cos u du$$       

 (Integrating by parts)

$$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$

$$I=\dfrac{\cos u e^{2u}}{2}-\displaystyle\int\dfrac{(-\sin u)e^{2u}}{2}du$$

$$=\dfrac{\cos u.e^{2u}}{2}+\dfrac{1}{2}\left[\dfrac{\sin u e^{2u}}{2}-\displaystyle\int\dfrac{\cos u e^{2u}}{2}du\right]$$

$$\Rightarrow I=\dfrac{e^{2u}\cos u}{2}+\dfrac{e^{2u}\sin u}{4}-\dfrac{I}{4}$$

$$\dfrac{5I}{4}=\dfrac{e^{2u}}{2}\left(\cos u+\dfrac{\sin u}{2}\right)+C$$

$$I=\dfrac{1}{5}\left[2\cos u+\sin u\right]e^{24}+C$$

$$I=\dfrac{1}{5}(2\sqrt{1-x^{2}}+x)e^{2\sin^{-1}(x)}+C$$
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