Integrate the following function with respect | vedclass

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

Question

(A) Let $I = \int \frac{\sin(	an^{-1} x)}{1+x^2} dx$. We know that the derivative of $	an^{-1} x$ is $\frac{1}{1+x^2}$. Let $t = 	an^{-1} x$. Then,

Vedclass · Accepted Answer

$-\cos(	an^{-1} x) + C$

Vedclass · Answer

(A) Let $I = \int \frac{\sin(	an^{-1} x)}{1+x^2} dx$. We know that the derivative of $	an^{-1} x$ is $\frac{1}{1+x^2}$. Let $t = 	an^{-1} x$. Then,$dt = \frac{1}{1+x^2} dx$. Substituting these into the integral,we get: $I = \int \sin(t) dt$. Integrating $\sin(t)$ with respect to $t$,we get: $I = -\cos(t) + C$. Substituting back $t = 	an^{-1} x$,we get: $I = -\cos(	an^{-1} x) + C$.

Integrate the following function with respect to $x$: $\frac{\sin(\tan^{-1} x)}{1+x^2}$

Similar Questions

$\int \frac{x}{\sqrt{x+4}} \, dx = $ . . . . . . $+ C, x > -4$.

$\int \cos ^{3} x \cdot e^{\log (\sin x)} d x$ is equal to

$\int \sqrt{\operatorname{cosec} x - \sin x} \, dx$ is equal to

If $f(x) = \int \frac{\sin 2x + 2 \cos x}{4 \sin^2 x + 5 \sin x + 1} \, dx$ and $f(0) = 0$, then $f\left(\frac{\pi}{6}\right) =$

$\int \frac{dx}{\sqrt{2x - x^2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module