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Question Number 125889 by bramlexs22 last updated on 15/Dec/20

$$\int \frac{\tan {}^{3}x}{\sqrt{\sec x}}dx?$$

Answered by bobhans last updated on 15/Dec/20

$$\int \frac{\tan x(\sec {}^{2}x-1)}{\sqrt{\sec x}}dx=$$$$[\sec x=u^2\Rightarrow \tan xdx=\frac{2du}{u}]$$$$I=\int \frac{(u^4-1)}{u}\left(\frac{2du}{u}\right)=2\int (u^2-u^{-2})du$$$$=2(\frac{1}{3}{u}^3+\frac{1}{u})+c$$$$=\frac{2\sqrt{\sec {}^{3}x}}{3}+\frac{2}{\sqrt{\sec x}}+c$$

Answered by Dwaipayan Shikari last updated on 15/Dec/20

$$\int \frac{{tanxsec}^{2}x}{\sqrt{secx}}-\frac{tanx}{\sqrt{secx}}dxsecx=t^2\Rightarrow secxtanx=\frac{dt}{dx}$$$$=2\int t^{2}dt-\int \frac{1}{t^2}=\frac{2}{3}{t}^3+\frac{2}{t}=\frac{2}{3}\sqrt{{sec}^{3}x}+\frac{2}{\sqrt{secx}}+C$$

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