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Question Number 115896 by bemath last updated on 29/Sep/20

$$\int \frac{\sec {}^{4}xdx}{\sqrt{\tan {}^{3}x}}=?$$

Answered by TANMAY PANACEA last updated on 29/Sep/20

$$\int \frac{(1+t^2)dt}{t^{\frac{3}{2}}}[t=tanx\frac{dt}{dx}={sec}^{2}x]$$$$\int t^{-\frac{3}{2}}+t^{\frac{1}{2}}dt$$$$=\frac{t^{-\frac{1}{2}}}{\frac{-1}{2}}+\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c$$$$=-2{\left(tanx\right)}^{\frac{-1}{2}}+\frac{2}{3}{\left(tanx\right)}^{\frac{3}{2}}+C$$$$$$

Commented by bemath last updated on 29/Sep/20

$$thankyousir$$

Answered by Ar Brandon last updated on 29/Sep/20

$$\mathcal{I}=\int \frac{{\sec }^{4}x}{\sqrt{{\tan }^{3}x}}\mathrm{d}x=\int \frac{{\sec }^{2}x}{\sqrt{{\tan }^{3}x}}\mathrm{d}\left(\tan x\right)$$$$=\int \frac{1+{\tan }^{2}x}{\sqrt{{\tan }^{3}x}}\mathrm{d}\left(\tan x\right)=\int \{\frac{1}{\sqrt{{\tan }^{3}x}}+\frac{{\tan }^{2}x}{\sqrt{{\tan }^{3}x}}\}\mathrm{d}\left(\tan x\right)$$$$=\int \{{\left(\tan x\right)}^{-\frac{3}{2}}+{\left(\tan x\right)}_{}^{\frac{1}{2}}\}\mathrm{d}\left(\tan x\right)$$$$=-\frac{2}{\sqrt{\tan x}}+\frac{2\sqrt{{\tan }^{3}x}}{3}+\mathcal{C}$$

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