int {frac{{sec x cdot csc x}}{{2cot x - sec x | vedclass

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Question

(A) Let $I = \int {\frac{{\sec x \cdot \csc x}}{{2\cot x - \sec x \cdot \csc x}}} dx$. Expressing the integrand in terms of $\sin x$ and $\cos x$: $\s

Vedclass · Accepted Answer

$\frac{1}{2}\ln |\sec 2x + 	an 2x| + c$

Vedclass · Answer

(A) Let $I = \int {\frac{{\sec x \cdot \csc x}}{{2\cot x - \sec x \cdot \csc x}}} dx$. Expressing the integrand in terms of $\sin x$ and $\cos x$: $\sec x \cdot \csc x = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$. $2\cot x - \sec x \cdot \csc x = \frac{2\cos x}{\sin x} - \frac{1}{\sin x \cos x} = \frac{2\cos^2 x - 1}{\sin x \cos x} = \frac{\cos 2x}{\sin x \cos x}$. Substituting these into the integral: $I = \int \frac{1/(\sin x \cos x)}{\cos 2x / (\sin x \cos x)} dx = \int \frac{1}{\cos 2x} dx = \int \sec 2x dx$. Using the standard integral $\int \sec u du = \ln |\sec u + 	an u| + c$,we get: $I = \frac{1}{2} \ln |\sec 2x + 	an 2x| + c$.

$\int {\frac{{\sec x \cdot \csc x}}{{2\cot x - \sec x \cdot \csc x}}} dx$ is equal to (where $c$ is the constant of integration).

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