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Question Number 119038 by MJS_new last updated on 21/Oct/20

$$\underset{-\pi /4}{\stackrel{+\pi /4}{\int }}\frac{\sqrt{1+\tan x}}{\sqrt{1-\tan x}}dx$$

Answered by mindispower last updated on 21/Oct/20

$$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\sqrt{1-tg\left(x\right)}}{\sqrt{1+tg\left(x\right)}}dx,I=\int \frac{\sqrt{1+tg\left(x\right)}}{\sqrt{1-tg\left(x\right)}}dx$$$$2I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\frac{\sqrt{1-tg\left(x\right)}}{\sqrt{1+tg\left(x\right)}}+\frac{\sqrt{1+tg\left(x\right)}}{\sqrt{1-tg\left(x\right)}})dx$$$$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1-tg\left(x\right)+1+tg\left(x\right)}{\sqrt{1-{tg}^2\left(x\right)}}dx=2\int \frac{dx}{\sqrt{1-{tg}^2\left(x\right)}}$$$$I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{\sqrt{1-{tg}^2\left(x\right)}}=2{\int }_0^{\frac{\pi }{4}}\frac{dx}{\sqrt{1-{tg}^2\left(x\right)}}$$$$lettg\left(x\right)=t\Rightarrow x=arctan\left(t\right)\Rightarrow dx=\frac{dt}{1+t^2}$$$$\Rightarrow 2{\int }_0^1\frac{dt}{\sqrt{1-t^2}(1+t^2)}$$$$t=sin\left(w\right)\Rightarrow dt=cos\left(w\right),\sqrt{1-t^2}=cos\left(w\right)$$$$\iff 2{\int }_0^{\frac{\pi }{2}}\frac{dw}{1+{sin}^2\left(w\right)}=2\int \frac{dw}{1+{cos}^2\left(w\right)}$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{1}{{cos}^2\left(w\right)(2+{tg}^2\left(w\right))}=2{\int }_0^{\frac{\pi }{2}}\frac{d\left(tg\left(w\right)\right)}{2+{tg}^2\left(w\right)}$$$$={\int }_0^{\frac{\pi }{2}}\frac{d\left(tg\left(w\right)\right)}{1+{\left(\frac{tg\left(w\right)}{\sqrt{2}}\right)}^2}=\sqrt{2}{\left[arctan\left(\frac{tg\left(w\right)}{\sqrt{2}}\right)\right]}_0^{\frac{\pi }{2}}$$$$\frac{\pi }{2}.\sqrt{2}=\frac{\pi }{\sqrt{2}},maybeesonmistacks$$

Commented by MJS_new last updated on 21/Oct/20

$$\mathrm{thank}\mathrm{you},\mathrm{I}\mathrm{get}\mathrm{the}\mathrm{same}\mathrm{answer}$$

Commented by mindispower last updated on 21/Oct/20

$$withepleasursir$$

Commented by $@y@m last updated on 22/Oct/20

@mindispower. Please expain me first line of the solution. How signs of Numerator and Denominator got interchanged?

Commented by Dwaipayan Shikari last updated on 22/Oct/20

$${\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\sqrt{\frac{1-tanx}{1+tanx}}={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\sqrt{\frac{1+tanx}{1-tanx}}$$$$Because$$$${\int }_q^{p}f\left(x\right)dx={\int }_q^{p}f(p+q-x)dx$$

Commented by $@y@m last updated on 22/Oct/20

@Dwaipayan অনেক ধন্যবাদ. আমি ভুলে গেছিলাম.

Commented by Dwaipayan Shikari last updated on 22/Oct/20

ধন্যবাদ। আপনি কোথায় থাকেন?

Commented by $@y@m last updated on 22/Oct/20

রাঁচি

Commented by MJS_new last updated on 22/Oct/20

Beautiful letters! Sadly I cannot read them.

Commented by Dwaipayan Shikari last updated on 22/Oct/20

first comment Many thanks I have forgotten second comment Thanking you. Where do you live? third comment Ranchi fourth comment সুন্দর হরফ! দু:খিত আমি এটা পড়তে পাচ্ছি না

Answered by $@y@m last updated on 22/Oct/20

$$\int \frac{\sqrt{1+\tan x}}{\sqrt{1-\tan x}}dx$$$$=\int \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx$$$$=\int \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}\times \frac{\cos x+\sin x}{\cos x+\sin x}}dx$$$$=\int \sqrt{\frac{{(\cos x+\sin x)}^2}{{\cos }^{2}x-{\sin }^{2}x}}dx$$$$=\int \frac{\cos x+\sin x}{\sqrt{(}{\cos }^{2}x-{\sin }^{2}x)}dx$$$$=\int \frac{\cos x}{\sqrt{1-2\sin {}^{2}x}}dx+\int \frac{\sin x}{\sqrt{2\cos {}^{2}x-1}}dx$$$$=\underset{-\frac{1}{\sqrt{2}}}{\stackrel{\frac{1}{\sqrt{2}}}{\int }}\frac{dp}{\sqrt{1-2p^2}}-\underset{\frac{1}{\sqrt{2}}}{\stackrel{\frac{1}{\sqrt{2}}}{\int }}\frac{dq}{\sqrt{2q^2-1}}$$$$\text{Missing left or extra right}$$$$=\frac{1}{\sqrt{2}}\{{\sin }^{-1}\left(1\right)-{\sin }^{-1}(-1)\}$$$$=\frac{1}{\sqrt{2}}(\frac{\pi }{2}-\frac{-\pi }{2})$$$$=\frac{\pi }{\sqrt{2}}$$

Commented by MJS_new last updated on 22/Oct/20

$$\mathrm{thank}\mathrm{you}$$

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