int frac{1}{1 + cos^2 x} dx = | vedclass

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Question

(C) To evaluate the integral $I = \int \frac{1}{1 + \cos^2 x} dx$,divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x}{\sec^2

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} 	an^{-1}\left( \frac{1}{\sqrt{2}} 	an x ight) + c$

Vedclass · Answer

(C) To evaluate the integral $I = \int \frac{1}{1 + \cos^2 x} dx$,divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x}{\sec^2 x + 1} dx$ Since $\sec^2 x = 1 + 	an^2 x$,we have: $I = \int \frac{\sec^2 x}{1 + 	an^2 x + 1} dx = \int \frac{\sec^2 x}{	an^2 x + 2} dx$ Let $t = 	an x$,then $dt = \sec^2 x dx$: $I = \int \frac{dt}{t^2 + 2} = \int \frac{dt}{t^2 + (\sqrt{2})^2}$ Using the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$: $I = \frac{1}{\sqrt{2}} 	an^{-1}\left( \frac{t}{\sqrt{2}} ight) + c$ Substituting $t = 	an x$ back: $I = \frac{1}{\sqrt{2}} 	an^{-1}\left( \frac{1}{\sqrt{2}} 	an x ight) + c$.

$\int \frac{1}{1 + \cos^2 x} dx = $

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