int {{e^x}left[ {{{sin }^{ - 1}}frac{x}{a} + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We use the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Let $f(x) = \sin^{-1}(\frac{x}{a})$. Then,the derivative $f'(x)

Vedclass · Accepted Answer

${e^x}{\sin ^{ - 1}}\frac{x}{a} + c$

Vedclass · Answer

(C) We use the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Let $f(x) = \sin^{-1}(\frac{x}{a})$. Then,the derivative $f'(x) = \frac{d}{dx}(\sin^{-1}(\frac{x}{a})) = \frac{1}{\sqrt{1 - (x/a)^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \cdot \frac{1}{a} = \frac{a}{\sqrt{a^2 - x^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{a^2 - x^2}}$. Since the integrand is of the form $e^x [f(x) + f'(x)]$,the integral is $e^x f(x) + c$. Therefore,$\int e^x [\sin^{-1}(\frac{x}{a}) + \frac{1}{\sqrt{a^2 - x^2}}] dx = e^x \sin^{-1}(\frac{x}{a}) + c$.

$\int {{e^x}\left[ {{{\sin }^{ - 1}}\frac{x}{a} + \frac{1}{{\sqrt {{a^2} - {x^2}} }}} \right]dx = }$

Similar Questions

$\int e^{\tan ^{-1} x} \left(1 + \frac{x}{1 + x^2} \right) dx$

$\int e^{\tan x}(\sec ^{2} x+\sec ^{3} x \sin x) d x$ is equal to

If $I = \int e^{\sin \theta} (\log \sin \theta + \operatorname{cosec}^2 \theta) \cos \theta \, d\theta$,then $I$ is equal to

$\int {\left\{ \frac{\log x - 1}{1 + (\log x)^2} \right\}}^2 dx$ is equal to

The value of $\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x+\cot x) d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module