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Question Number 122940 by CanovasCamiseros last updated on 21/Nov/20

Commented by CanovasCamiseros last updated on 21/Nov/20

$$\mathit{h}\mathit{o}\mathit{w}\mathit{c}\mathit{a}\mathit{n}\mathit{i}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{i}\mathit{t}?$$

Commented by mr W last updated on 21/Nov/20

$$gotoQ73012$$

Answered by TANMAY PANACEA last updated on 21/Nov/20

$$t^2=tanx\to 2t\frac{dt}{dx}={sec}^{2}x$$$$dx=\frac{2t}{1+t^4}dt$$$$\int \frac{t\times 2tdt}{1+t^4}$$$$\int \frac{2dt}{t^2+\frac{1}{t^2}}$$$$\int \frac{1-\frac{1}{t^2}+1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$$$$\int \frac{d(t+\frac{1}{t})}{{(t+\frac{1}{t})}^2-2}+\int \frac{d(t-\frac{1}{t})}{{(t-\frac{1}{t})}^2+2}$$$$useformula\int \frac{dx}{x^2-a^2}\mathrm{\&}\frac{\int dx}{x^2+a^2}$$$$\frac{1}{2\sqrt{2}}ln\left(\frac{t+\frac{1}{t})-\sqrt{2}}{(t+\frac{1}{t}+\sqrt{2}}\right)+\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)$$$$puyt=\sqrt{tanx}$$$$$$$$$$

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