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Question Number 108145 by bemath last updated on 15/Aug/20

$$\frac{⊚\mathcal{B}e\mathcal{M}ath⊚}{\mathrm{\Pi }}$$ $$\left(1\right)\cos {}^2\left(\frac{x}{4}\right)>\frac{\sqrt{2}}{2}+\sin {}^2\left(\frac{x}{4}\right)$$ $$\left(2\right)\int \sqrt{\frac{1+x}{1-x}}dx$$ $$\left(3\right)\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sqrt{\tan x}}{{(\cos x+\sin x)}^2}dx$$

Commented bybemath last updated on 15/Aug/20

dear admint tinkutara. i have used v.2.138. problem for which i can not edit answer

Commented byTinku Tara last updated on 15/Aug/20

$$\mathrm{Just}\mathrm{tested}.\mathrm{What}\mathrm{error}\mathrm{message}$$ $$\mathrm{do}\mathrm{u}\mathrm{get}?$$

Answered by 1549442205PVT last updated on 15/Aug/20

$$1){\cos }^2\frac{\mathrm{x}}{4}>\frac{\sqrt{2}}{2}+{\sin }^2\frac{\mathrm{x}}{4}\iff {\cos }^2\frac{\mathrm{x}}{4}-{\sin }^2\frac{\mathrm{x}}{4}>\frac{\sqrt{2}}{2}$$ $$\iff \cos \frac{\mathrm{x}}{2}>\frac{\sqrt{2}}{2}=\cos \frac{\pi }{4}$$ $$\iff 2\mathbf{k}\mathit{\pi }-\frac{\mathit{\pi }}{4}<\mathbf{x}<\frac{\mathit{\pi }}{4}+2\mathbf{k}\mathit{\pi }$$ $$2)\mathrm{Put}\mathrm{x}=\cos \theta \Rightarrow \mathrm{dx}=-\sin \theta \mathrm{d}\theta$$ $$\mathrm{I}=-\int \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}\sin \theta =-\int \sqrt{\frac{{2\cos }^2\frac{\theta }{2}}{{2\sin }^2\frac{\theta }{2}}}\times \frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\mathrm{d}\theta$$ $$=-\int \cot \frac{\theta }{2}\times \frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\mathrm{d}\theta =-2\int \frac{\mathrm{d}\theta }{1+{\tan }^2\frac{\theta }{2}}$$ $$=-2\int {\cos }^2\frac{\theta }{2}\mathrm{d}\theta =-\int (1+\cos \theta )\mathrm{d}\theta$$ $$=-\mathit{\theta }-\mathbf{sin}\mathit{\theta }=-{\mathbf{sin}}^{-1}\mathbf{x}-\sqrt{1-{\mathbf{x}}^2}+\mathbf{C}$$ $$3){\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\mathrm{tanx}}}{{(\mathrm{cosx}+\mathrm{sinx})}^2}\mathrm{dx}=\int \frac{\sqrt{\mathrm{tanx}}}{1+2\sin 2\mathrm{x}}\mathrm{dx}$$ $$={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\mathrm{tanx}}}{1+\frac{2\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}}\mathrm{dx}.\mathrm{Put}\mathrm{tanx}=\mathrm{t}$$ $$\Rightarrow \mathrm{dt}=(1+{\mathrm{t}}^2)\mathrm{dx},\frac{\sqrt{\mathrm{tanx}}}{1+\frac{2\mathrm{tanx}}{1+{\tan }^2\mathrm{x}}}=\frac{(1+{\mathrm{t}}^2)\sqrt{\mathrm{t}}}{{(1+\mathrm{t})}^2}$$ $$\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{(1+{\mathrm{t}}^2)}^2\sqrt{\mathrm{t}}}{{(1+\mathrm{t})}^2}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{{[{(1+\mathrm{t})}^2-2\mathrm{t}]}^2\sqrt{\mathrm{t}}}{{(1+\mathrm{t})}^2}$$ $$={\int }_0^{\mathrm{\infty }}[{(1+\mathrm{t})}^2-4\mathrm{t}+\frac{{4\mathrm{t}}^2}{{(1+\mathrm{t})}^2}]\sqrt{\mathrm{t}}\mathrm{dt}$$ $$={\int }_0^{\mathrm{\infty }}[(1-2\mathrm{t}+{\mathrm{t}}^2)\sqrt{\mathrm{t}}+\frac{{4\mathrm{t}}^2\sqrt{\mathrm{t}}}{{(1+\mathrm{t})}^2}]\mathrm{dt}$$ $$=(\frac{2}{3}{\mathrm{t}}^{3/2}-\frac{4}{5}{\mathrm{t}}^{5/2}+\frac{2}{7}{\mathrm{t}}^{7/2})+{\int }_0^{\mathrm{\infty }}\frac{{4\mathrm{t}}^2\sqrt{\mathrm{t}}}{{(1+\mathrm{t})}^2}\mathrm{dt}$$ $$\mathrm{Put}\sqrt{\mathrm{t}}=\mathrm{u}\Rightarrow \mathrm{du}=\left(1/2\sqrt{\mathrm{t}}\right)\mathrm{dt}\Rightarrow \mathrm{dt}=2\mathrm{udu}$$ $$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{{4\mathrm{t}}^2\sqrt{\mathrm{t}}}{{(1+\mathrm{t})}^2}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{{8\mathrm{u}}^6}{{(1+{\mathrm{u}}^2)}^2}\mathrm{du}.$$ $${\mathrm{u}}^6={(1+{\mathrm{u}}^2)}^3-{3\mathrm{u}}^2(1+{\mathrm{u}}^2)-1={(1+{\mathrm{u}}^2)}^3$$ $$-3{(1+{\mathrm{u}}^2)}^2+3(1+{\mathrm{u}}^2)-1.\mathrm{Hence},$$ $$\mathrm{A}=8{\int }_0^{\mathrm{\infty }}[(1+{\mathrm{u}}^2)-3+\frac{3}{(1+{\mathrm{u}}^2)}-\frac{1}{{(1+{\mathrm{u}}^2)}^2}]\mathrm{du}$$ $$=8(\frac{{\mathrm{u}}^3}{3}-2\mathrm{u}+{3\tan }^{-1}\left(\mathrm{u}\right))\mid -{\int }_0^{\mathrm{\infty }}\frac{8}{{(1+{\mathrm{u}}^2)}^2}\mathrm{du}$$ $$\mathrm{Put}\mathrm{u}=\tan \theta \Rightarrow \mathrm{du}=(1+{\mathrm{u}}^2)\mathrm{d}\theta$$ $$\mathrm{B}={\int }_0^{\frac{\pi }{2}}\frac{8}{1+{\tan }^2\theta }\mathrm{d}\theta =8{\int }_0^{\frac{\pi }{2}}{\cos }^2\theta \mathrm{d}\theta$$ $$=4{\int }_0^{\frac{\pi }{2}}(1+\cos 2\theta )\mathrm{d}\theta =4\theta +2\sin 2\theta {\mid }_0^{\frac{\pi }{2}}=\pi$$ $$\mathbf{The}\mathbf{value}\mathbf{of}\mathbf{this}\mathbf{integral}{\mathbf{isn}}^{\prime }\mathbf{t}\mathbf{defined}$$

Commented bymathdave last updated on 15/Aug/20

$$answerforquestion3is\frac{\pi }{2}$$

Commented bybemath last updated on 15/Aug/20

$$thankyouallsir$$

Answered by Dwaipayan Shikari last updated on 15/Aug/20

$$2)\int \sqrt{\frac{1+x}{1-x}}dx$$ $$\int \frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}dx$$ $${sin}^{-1}x-\frac{1}{2}\int \frac{-2x}{\sqrt{1-x^2}}dt$$ $${sin}^{-1}x-\sqrt{1-x^2}+C$$

Answered by john santu last updated on 15/Aug/20

$$\frac{\divideontimes JS\divideontimes }{\mathrm{♡}}$$ $$I=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sqrt{\tan x}}{\cos {}^{2}x{(1+\tan x)}^2}dx$$ $$I=\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sqrt{\tan x}\sec {}^{2}xdx}{{(1+\tan x)}^2}[\tan x=b]$$ $$I=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\sqrt{b}}{{(1+b)}^2}db[setb=\frac{q}{1-q}]$$ $$I=\underset{0}{\stackrel{1}{\int }}{q}^{1/2}{(1-q)}^{-1/2}dq$$ $$Recallbethafunction$$ $$\underset{0}{\stackrel{1}{\int }}{x}^{m-1}{(1-x)}^{n-1}dx=B(m,n)$$ $$=\frac{\mathrm{\Gamma }\left(m\right)\mathrm{\Gamma }\left(n\right)}{\mathrm{\Gamma }(m+n)}.henceweobtain$$ $$m=\frac{3}{2},n=\frac{1}{2}.$$ $$I=\frac{\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{1!}=\frac{1}{2}{\left(\sqrt{\pi }\right)}^2=\frac{\pi }{2}$$

Commented bymathdave last updated on 15/Aug/20

$$brilliantsolution$$

Commented bymohammad17 last updated on 15/Aug/20

$$sircanyouexactlyhow(m=\frac{3}{2},n=\frac{1}{2})$$

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