k in N, int frac{1-k cos ^2 x}{sin ^k x cdot | vedclass

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

Question

(B) Let $I = \int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x$. We can rewrite the integrand as: $I = \int \frac{\sec ^2 x - k}{\sin ^k x} d x

Vedclass · Accepted Answer

$\frac{	an x}{\sin ^k x}+C$

Vedclass · Answer

(B) Let $I = \int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x$. We can rewrite the integrand as: $I = \int \frac{\sec ^2 x - k}{\sin ^k x} d x = \int (\sin x)^{-k} \sec ^2 x d x - k \int \csc ^k x d x$. Using integration by parts on the first integral $\int (\sin x)^{-k} \sec ^2 x d x$,let $u = (\sin x)^{-k}$ and $dv = \sec ^2 x d x$. Then $du = -k(\sin x)^{-k-1} \cos x d x$ and $v = 	an x$. $I = (\sin x)^{-k} 	an x - \int 	an x \cdot (-k)(\sin x)^{-k-1} \cos x d x - k \int \csc ^k x d x$. Since $	an x \cdot \cos x = \sin x$,the integral becomes: $I = \frac{	an x}{\sin ^k x} + k \int (\sin x)^{-k-1} \sin x d x - k \int \csc ^k x d x$. $I = \frac{	an x}{\sin ^k x} + k \int (\sin x)^{-k} d x - k \int \csc ^k x d x$. $I = \frac{	an x}{\sin ^k x} + k \int \csc ^k x d x - k \int \csc ^k x d x + C$. $I = \frac{	an x}{\sin ^k x} + C$.

$k \in N, \int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x=$

Similar Questions

$\int \frac{d x}{\sqrt{\left(5+2 x+x^2\right)^3}}$ is equal to

If $\int \sin (101 x)(\sin x)^{99} d x=\frac{\sin (100 x)(\sin x)^\lambda}{\mu}+c$ then $\frac{\lambda}{\mu}=$

$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to

$\int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x=$

$\int \sec x \tan^3 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module