int frac{sin ^6 x}{cos ^8 x} d x= | vedclass

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Question

(B) We are given the integral $I = \int \frac{\sin ^6 x}{\cos ^8 x} d x$. Rewrite the integrand as: $I = \int \frac{\sin ^6 x}{\cos ^6 x} \cdot \frac{

Vedclass · Accepted Answer

$\frac{	an^7 x}{7}+c$

Vedclass · Answer

(B) We are given the integral $I = \int \frac{\sin ^6 x}{\cos ^8 x} d x$. Rewrite the integrand as: $I = \int \frac{\sin ^6 x}{\cos ^6 x} \cdot \frac{1}{\cos ^2 x} d x$. Since $\frac{\sin x}{\cos x} = 	an x$ and $\frac{1}{\cos x} = \sec x$,we have: $I = \int 	an ^6 x \sec ^2 x d x$. Let $u = 	an x$,then $du = \sec ^2 x d x$. Substituting these into the integral,we get: $I = \int u ^6 d u = \frac{u ^7}{7} + c$. Substituting back $u = 	an x$,we obtain: $I = \frac{	an ^7 x}{7} + c$.

$\int \frac{\sin ^6 x}{\cos ^8 x} d x=$

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