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

To evaluate $\int x^3 e^{3x^2 + 5} dx$,the simplest way is to

Find $\int \frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}} dx$

The value of $\int \frac{e^{x}(1+x) dx}{\cos^{2}(x e^{x})}$ is equal to

Find the integral of the function $\sin ^{3} x \cos ^{3} x$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module