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Question Number 211727 by Spillover last updated on 18/Sep/24

Answered by MrGaster last updated on 03/Nov/24

$$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{(\pi -4x)\tan x(1+\tan x)}{1-{\tan }^{2}x}dx$$$$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{(\pi -4x)\tan x+(\pi -4x){\tan }^{2}x}{{\cos }^{2}x}dx$$$$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\pi -4x)\tan x{\sec }^{2}xdx+{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\pi -4x){\tan }^{2}x{\sec }^{2}xdx$$$$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\pi -4x)\tan xd\left(\tan x\right)+{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\pi -4x){\tan }^{2}xd\left(\tan x\right)$$$$={\left[\frac{(\pi -4x){\tan }^{2}x}{2}\right]}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}-{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(-4)\frac{{\tan }^{2}x}{2}dx+{\left[\frac{(\pi -4x){\tan }^{3}x}{3}\right]}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}-{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(-4)\frac{{\tan }^{3}x}{3}dx$$$$={[\frac{\pi {\tan }^{2}x}{2}-2{\tan }^{2}x]}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}+{[\frac{\pi {\tan }^{3}x}{3}-\frac{4{\tan }^{3}x}{3}]}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}$$$$=[\frac{\pi }{2}-2]{\tan }^2\frac{\pi }{4}-[\frac{\pi }{2}-2]{\tan }^2-\frac{\pi }{4}+[\frac{\pi }{3}-\frac{4}{3}]{\tan }^3\frac{\pi }{4}-[\frac{\pi }{3}-\frac{4}{3}]{\tan }^3-\frac{\pi }{4}$$$$=[\frac{\pi }{2}-2]-[\frac{\pi }{2}-2]+[\frac{\pi }{3}-\frac{4}{3}]-[\frac{\pi }{3}-\frac{4}{3}]$$$$=0+0$$$$=0$$

Commented by Spillover last updated on 03/Nov/24

$$great.thankyou$$

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