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Question Number 187683 by 073 last updated on 20/Feb/23

Answered by anurup last updated on 20/Feb/23

$${\mathrm{I}}_1=\int \sqrt{\tan x}dx$$$$=\frac{1}{2}\int \{(\sqrt{\tan x}+\sqrt{\cot x})+(\sqrt{\tan x}-\sqrt{\cot x})\}dx$$$$=\frac{1}{2}\int \{(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}})+(\sqrt{\frac{\sin x}{\cos x}}-\sqrt{\frac{\cos x}{\sin x}})\}dx$$$$=\frac{1}{2}\int \{\left(\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\right)+\left(\frac{\sin x-\cos x}{\sqrt{\sin x\cos x}}\right)\}dx$$$$=\frac{1}{\sqrt{2}}\int \{\left(\frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}\right)+\left(\frac{\sin x-\cos x}{\sqrt{2\sin x\cos x}}\right)\}dx$$$$=\frac{1}{\sqrt{2}}\int \{\left(\frac{\sin x+\cos x}{\sqrt{1-(1-2\sin x\cos x)}}\right)+\left(\frac{\sin x-\cos x}{\sqrt{(1+2\sin x\cos x)-1}}\right)\}dx$$$$=\frac{1}{\sqrt{2}}\int \{\left(\frac{\sin x+\cos x}{\sqrt{1-{(\sin x-\cos x)}^2}}\right)-\left(\frac{\cos x-\sin x}{\sqrt{{(\sin x+\cos x)}^2-1}}\right)\}dx$$$$=\frac{1}{\sqrt{2}}\int \frac{du}{\sqrt{1-u^2}}-\frac{1}{\sqrt{2}}\int \frac{dv}{\sqrt{v^2-1}}[u=\sin x-\cos x,v=\sin x+\cos x]$$$$=\frac{1}{\sqrt{2}}{\sin }^{-1}\left(u\right)-\frac{1}{\sqrt{2}}\ln \mid v+\sqrt{v^2-1}\mid +{\mathrm{C}}_1$$$$=\frac{1}{\sqrt{2}}{\sin }^{-1}(\sin x-\cos x)-\frac{1}{\sqrt{2}}\ln \mid \sin x+\cos x+\sqrt{\sin 2x}\mid +{\mathrm{C}}_1$$$${\mathrm{I}}_2=\int \frac{\sqrt[3]{3{xe}^{x^2}}}{1+x^7}dx$$$$=\int \frac{{\left(3x\right)}^{\frac{1}{3}}{e}^{\frac{x^2}{3}}}{1+x^7}dx$$$$$$

Commented by anurup last updated on 20/Feb/23

$${\mathrm{I}}_2\mathrm{not}\mathrm{done}$$

Commented by 073 last updated on 21/Feb/23

$$\mathrm{thanks}\mathrm{alot}$$

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