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Question Number 197396 by sonukgindia last updated on 16/Sep/23

Commented by Frix last updated on 16/Sep/23

$$\mathrm{We}\mathrm{had}\mathrm{this}\mathrm{several}\mathrm{times}\mathrm{before}.$$$$t=\sqrt{\tan x}\Rightarrow 2\int \frac{t^2}{t^4+1}dt$$$$\mathrm{which}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{by}\mathrm{decomposing}$$

Answered by universe last updated on 16/Sep/23

$$2I=\int \underset{I_1}{\underbrace{(\sqrt{\tan x}+\sqrt{\cot x})}dx}+\underset{I_2}{\underbrace{\int (\sqrt{\tan x}-\sqrt{\cot x})dx}}$$$$I_1=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}dx$$$$I_1=\sqrt{2}\int \frac{(\sin x+\cos x)}{\sqrt{1-{(\sin x-\cos x)}^2}}dx$$$$let$$$$\sin x-\cos x=t\Rightarrow (\cos x+\sin x)dx=dt$$$$I_1=\sqrt{2}\int \frac{dt}{\sqrt{1-t^2}}=\sqrt{2}{\sin }^{-1}t+C_1$$$$I_1=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C_1$$$$now$$$$I_2=-\sqrt{2}\int \frac{(-\sin x+\cos x)}{\sqrt{{(\sin x+\cos x)}^2-1}}dx$$$$let\sin x+\cos x=y$$$$(\cos x-\sin x)dx=dy$$$$I_2=-\sqrt{2}\int \frac{dy}{\sqrt{y^2-1}}$$$$I_2=-\sqrt{2}{\cosh }^{-1}y+C_2$$$$I_2=-\sqrt{2}{\cosh }^{-1}(\sin x+\cos x)+C_2$$$$2I=I_1+I_2$$$$I=\frac{1}{\sqrt{2}}[{\sin }^{-1}(\sin x-\cos x)-{\cosh }^{-1}(\sin x+\cos x)+C$$

Commented by MM42 last updated on 16/Sep/23

$$\cancel{⋚}$$

Commented by universe last updated on 16/Sep/23

$$\underset{―}{\underbrace{⋚}}$$

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