Evaluate the integral int frac{x^2 + cos^2 x} | vedclass

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Question

(A) We are given the integral $I = \int \left( \frac{x^2 + \cos^2 x}{1 + x^2} \right) \operatorname{cosec}^2 x \, dx$. Rewrite the numerator as $(x^2

Vedclass · Accepted Answer

$-	an^{-1} x - \cot x + c$

Vedclass · Answer

(A) We are given the integral $I = \int \left( \frac{x^2 + \cos^2 x}{1 + x^2} ight) \operatorname{cosec}^2 x \, dx$. Rewrite the numerator as $(x^2 + 1 - 1 + \cos^2 x)$: $I = \int \left( \frac{x^2 + 1}{1 + x^2} - \frac{1}{1 + x^2} + \frac{\cos^2 x}{1 + x^2} ight) \operatorname{cosec}^2 x \, dx$ Simplify the expression: $I = \int \left( 1 - \frac{1}{1 + x^2} + \frac{\cos^2 x}{1 + x^2} ight) \operatorname{cosec}^2 x \, dx$ Distribute $\operatorname{cosec}^2 x$: $I = \int \left( \operatorname{cosec}^2 x - \frac{\operatorname{cosec}^2 x}{1 + x^2} + \frac{\cos^2 x}{(1 + x^2) \sin^2 x} ight) \, dx$ Since $\frac{\cos^2 x}{\sin^2 x} = \cot^2 x$: $I = \int \left( \operatorname{cosec}^2 x + \frac{\cot^2 x - \operatorname{cosec}^2 x}{1 + x^2} ight) \, dx$ Using the identity $\cot^2 x - \operatorname{cosec}^2 x = -1$: $I = \int \left( \operatorname{cosec}^2 x - \frac{1}{1 + x^2} ight) \, dx$ Integrating term by term: $I = -\cot x - 	an^{-1} x + c$.

Evaluate the integral $\int \frac{x^2 + \cos^2 x}{1 + x^2} \operatorname{cosec}^2 x \, dx$,where $c$ is the constant of integration.

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