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# Evaluate the definite integrals $\int\limits_0^{\pi}\large\frac{x\tan x}{\sec x+\tan x}$$\;dx ## 1 Answer Comment A) Need homework help? Click here. Toolbox: • (ii) \int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx Step 1: Given I=\int\limits_0^{\pi}\large\frac{x\tan x}{\sec x+\tan x}\;$$dx$
By applying the property $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$I=\int\limits_0^\pi\large\frac{(\pi-x)\tan (\pi-x)}{\sec(\pi-x)+\tan(\pi-x)}$$dx-----(1) I=\int\limits_0^\pi\large\frac{(\pi-x)\tan x}{\sec x+\tan x}$$dx$------(2)
$\tan(\pi-x)=-\tan x$
$\sec(\pi-x)--\sec x$
Step 2:
On adding equ(1) and equ(2) we get,
$2I=\int\limits_0^\pi\large\frac{\pi\tan x}{\sec x+\tan x}$$dx \;\;\;=\pi\int_0^\pi\large\frac{\tan x}{\sec x+\tan x}$$dx$
$\;\;\;=\pi\int_0^\pi\large\frac{\sin x/\cos x}{1/\cos x+\sin x/\cos x}$$dx \;\;\;=\pi\int_0^\pi\large\frac{\sin x}{1+\sin x}$$dx$
$\;\;\;=\pi\int_0^\pi\large\frac{\sin x(1-\sin x)}{1-\sin^2 x}$$dx Step 3: \sin^2 x+\cos^2 x=1 \Rightarrow 1-\sin^2 x=\cos^2 x \;\;\;=\pi\int_0^\pi\large\frac{\sin x-\sin ^2x)}{\cos^2x}$$dx$
$\;\;\;=\pi\int_0^\pi[\large\frac{\sin x}{\cos^2x}-\frac{\sin^2 x}{\cos^2 x}]$$dx \;\;\;=\pi\int\limits_0^\pi(\sec x\tan x-(\sec^2 x-1) dx 1+\tan^2 x=\sec^2 x 2I=\pi[\sec x-\tan x+x]_0^\pi \;\;\;\;=\pi[\sec \pi-\tan \pi+\pi -(\sec 0-\tan 0+0)] \;\;\;\;=\pi(-1+\pi-1) I=\large\frac{\pi}{2}$$(\pi-2)$