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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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If $f(a+b-x)=f(x)$ then $\int_{a}^{b}xf\left ( x \right )dx$ equals
• A. $\displaystyle \frac{b-a}{2}\int_{a}^{b}f\left ( x \right )dx$
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• C. $\displaystyle \frac{a+b}{2}\int_{a}^{b}f\left ( b-x \right )dx$
• D. $\displaystyle \frac{a+b}{2}\int_{a}^{b}f\left ( x \right )dx$

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