int frac{cos ^{n-1} x}{sin ^{n+1} x} d x (whe | vedclass

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

Vedclass · Accepted Answer

$\frac{-\cot ^{n} x}{n}+C$

Vedclass · Answer

(C) Let $I = \int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} d x$. We can rewrite the integrand as: $I = \int \frac{\cos ^{n-1} x}{\sin ^{n-1} x \cdot \sin^2 x} d x = \int \cot^{n-1} x \cdot \csc^2 x d x$. Let $t = \cot x$. Then $dt = -\csc^2 x d x$,which implies $\csc^2 x d x = -dt$. Substituting these into the integral: $I = \int t^{n-1} (-dt) = -\int t^{n-1} dt$. Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$: $I = -\frac{t^n}{n} + C$. Substituting $t = \cot x$ back: $I = -\frac{\cot^n x}{n} + C$.

$\int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} d x$ (where,$n \neq 0$) is equal to

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