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

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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: $I = \frac{	an x}{\sin ^k x} + k \int (\sin x)^{-k-1} \sin x d x - k \int \csc ^k x d x$. $I = \frac{	an x}{\sin ^k x} + k \int (\sin x)^{-k} d x - k \int \csc ^k x d x$. $I = \frac{	an x}{\sin ^k x} + k \int \csc ^k x d x - k \int \csc ^k x d x + C$. $I = \frac{	an x}{\sin ^k x} + C$.

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

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