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Question Number 161537 by cortano last updated on 19/Dec/21

$${\int }_0^{\frac{\pi }{4}}\frac{1+\tan {}^4\left(x\right)}{\cot {}^2\left(x\right)}dx=?$$

Answered by Ar Brandon last updated on 19/Dec/21

$$={\int }_0^{\frac{\pi }{4}}({\tan }^{2}x+{\tan }^{6}x)dx$$$$={\int }_0^{\frac{\pi }{4}}({\sec }^{2}x-1)dx+{\int }_0^{\frac{\pi }{4}}\left({\tan }^{4}x\right)({\sec }^{2}x-1)dx$$$$={[\tan x-x]}_0^{\frac{\pi }{4}}+{\left[\frac{{\tan }^{5}x}{5}\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\left({\tan }^{2}x\right)({\sec }^{2}x-1)dx$$$$=1-\frac{\pi }{4}+\frac{1}{5}-{\left[\frac{{\tan }^{3}x}{3}\right]}_0^{\frac{\pi }{4}}+{\int }_0^{\frac{\pi }{4}}({\sec }^{2}x-1)dx$$$$=\frac{6}{5}-\frac{\pi }{4}-\frac{1}{3}+{[\tan x-x]}_0^{\frac{\pi }{4}}=\frac{13}{15}-\frac{\pi }{4}+(1-\frac{\pi }{4})$$$$=\frac{28}{15}-\frac{\pi }{2}$$

Commented by Ar Brandon last updated on 19/Dec/21

$$1+{\tan }^{2}x={\sec }^{2}x$$$$\frac{d\left(\tan x\right)}{dx}={\sec }^{2}x$$

Commented by peter frank last updated on 20/Dec/21

$$\mathrm{good}$$

Answered by cortano last updated on 19/Dec/21

Commented by saboorhalimi last updated on 19/Dec/21

$$sirwhichsoftwaredidyouuse$$$$forwritingthissolution?$$

Commented by cortano last updated on 20/Dec/21

$$matheditorforpc$$

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