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Question Number 54616 by cesar.marval.larez@gmail.com last updated on 07/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19

$$a=tanxda={sec}^{2}xdx$$$$\int \frac{a^{2}da}{\sqrt{2+1+a^2}}$$$$\int \frac{a^2+3-3}{\sqrt{a^2+3}}da$$$$\int \sqrt{a^2+3}da-3\int \frac{da}{\sqrt{a^2+3}}$$$$nowuseformula$$$$\frac{a}{2}\sqrt{a^2+3}+\frac{3}{2}ln(a+\sqrt{a^2+3})-3ln(a+\sqrt{a^2+3})+c$$$$\frac{tanx}{2}\sqrt{{tan}^{2}x+3}-\frac{3}{2}ln(tanx+\sqrt{{tan}^{2}x+3})+c$$

Answered by MJS last updated on 08/Feb/19

$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{be}\mathrm{cheeky}$$$$\int \frac{{\sec }^{2}x{\tan }^{2}x}{\sqrt{2+{\sec }^{2}x}}dx=$$$$[t=\sqrt{2+{\sec }^{2}x}\to x=\arccos \left(\frac{1}{\sqrt{t^2-2}}\right);dx=\frac{{\sec }^{2}x\sqrt{2+{\sec }^{2}x}}{\tan x}dt]$$$$=\int \tan xdt=\int \sqrt{t^2-3}dt=$$$$[u=\frac{\sqrt{3}}{3}t\to dt=\sqrt{3}du]$$$$=3\int \sqrt{u^2-1}du=$$$$[v=\mathrm{arcosh}u\to du=\sinh vdv]$$$$=3\int {\sinh }^{2}vdv=\frac{3}{2}(v+\sinh v\cosh v)=$$$$=\frac{3}{2}(u\sqrt{u^2-1}-\mathrm{arcosh}u)=$$$$=\frac{1}{2}(t\sqrt{t^2-3}-3\mathrm{arcosh}\frac{t\sqrt{3}}{3})=$$$$=\frac{1}{2}\sqrt{{\sec }^{4}x+{\sec }^{2}x-2}-\frac{3}{2}\mathrm{arcosh}\sqrt{\frac{2+{\sec }^{2}x}{3}}+C$$

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