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nn 2 - 1
\int_{0}^{\frac{\pi }{2}}{\frac{sec^{2}x}{(secx+tanx)^{n}}}dx,n>1
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3 Answers
1\sec x +\tan x =t \\ \\ \\ \frac{dt}{t}=\sec x .dx\\ \\ \\ \int \frac{\sec x .dx }{(\sec x + \tan x)^n}=\int {t^{-n-1}dt}=\frac{-1}{nt^n}=-\frac{1}{n(\sec x + \tan x)^n} \\ \\ \\ \int_{0}^{\frac{\pi}{2}}{\sec x \frac{\sec x .dx }{(\sec x + \tan x)^n}}= \frac{-\cos ^ {n-1}x}{n(1 + \sin x )^n}|_0^{\frac{\pi}{2}}+\underline {\int_{0}^{\frac{\pi}{2}}{\sec x \tan x\frac{1}{n(\sec x + \tan x)^n}}}\\ \\ \\ \\ \text{now to evaluate this take tan as first function , then u will agin get I(n), which gives the answer }
1\int_{o}^{\frac{\pi}{2}}{\tan x \frac{\sec x}{\left( \tan x +\sec x\right)^n}}=-\sin x.\cos^{n-1}x\frac{1}{n(1+\sin x)^n}|_0^{\frac{\pi}{2}}+\frac{1}{n}\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2 x}{(\sec x +\tan x)^n}} \\ =0+\frac{I}{n}\\ \text{now put it in the equation } I=\frac{1}{n}+\frac{I}{n^2}\\ \boxed{I=\frac{n}{n^2-1} }