pi-4-pi-4-1-tan-x-1-tan-x-dx-

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

$$\underset{−\pi/\mathrm{4}} {\overset{+\pi/\mathrm{4}} {\int}}\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}{dx} \\ $$

Answered by mindispower last updated on 21/Oct/20

$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{1}−{tg}\left({x}\right)}}{\:\sqrt{\mathrm{1}+{tg}\left({x}\right)}}{dx},{I}=\int\frac{\sqrt{\mathrm{1}+{tg}\left({x}\right)}}{\:\sqrt{\mathrm{1}−{tg}\left({x}\right)}}{dx} \\ $$$$\mathrm{2}{I}=\int_{−\frac{\pi}{\mathrm{4}}\:} ^{\frac{\pi}{\mathrm{4}}} \left(\frac{\sqrt{\mathrm{1}−{tg}\left({x}\right)}}{\:\sqrt{\mathrm{1}+{tg}\left({x}\right)}}+\frac{\sqrt{\mathrm{1}+{tg}\left({x}\right)}}{\:\sqrt{\mathrm{1}−{tg}\left({x}\right)}}\right){dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}−{tg}\left({x}\right)+\mathrm{1}+{tg}\left({x}\right)}{\:\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \left({x}\right)}}{dx}=\mathrm{2}\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \left({x}\right)}} \\ $$$${I}=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \left({x}\right)}}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \left({x}\right)}} \\ $$$${let}\:{tg}\left({x}\right)={t}\Rightarrow{x}={arctan}\left({t}\right)\Rightarrow{dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$${t}={sin}\left({w}\right)\Rightarrow{dt}={cos}\left({w}\right),\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }={cos}\left({w}\right) \\ $$$$\Leftrightarrow\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dw}}{\mathrm{1}+{sin}^{\mathrm{2}} \left({w}\right)}=\mathrm{2}\int\frac{{dw}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({w}\right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({w}\right)\left(\mathrm{2}+{tg}^{\mathrm{2}} \left({w}\right)\right)}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left({tg}\left({w}\right)\right)}{\mathrm{2}+{tg}^{\mathrm{2}} \left({w}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left({tg}\left({w}\right)\right)}{\mathrm{1}+\left(\frac{{tg}\left({w}\right)}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}\left[{arctan}\left(\frac{{tg}\left({w}\right)}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\frac{\pi}{\mathrm{2}}.\sqrt{\mathrm{2}}=\frac{\pi}{\:\sqrt{\mathrm{2}}},{may}\:{bee}\:{son}\:{mistacks}\: \\ $$

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

$${withe}\:{pleasur}\:{sir} \\ $$

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

$$\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\mathrm{1}−{tanx}}{\mathrm{1}+{tanx}}}\:=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}} \\ $$$${Because} \\ $$$$\int_{{q}} ^{{p}} {f}\left({x}\right){dx}=\int_{{q}} ^{{p}} {f}\left({p}+{q}−{x}\right){dx} \\ $$

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

$$\int\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}{dx} \\ $$$$=\int\sqrt{\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}}{dx} \\ $$$$=\int\sqrt{\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}×\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}}{dx} \\ $$$$=\int\sqrt{\frac{\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{sin}^{\mathrm{2}} \:{x}}}{dx} \\ $$$$=\int\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\left.\:\sqrt{\left(\right.}\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)}{dx} \\ $$$$=\int\frac{\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}}}{dx}+\int\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}}}{dx} \\ $$$$=\underset{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{{dp}}{\:\sqrt{\mathrm{1}−\mathrm{2}{p}^{\mathrm{2}} }}−\underset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\int}}\frac{{dq}}{\:\sqrt{\mathrm{2}{q}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{p}\right)\underset{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\right]}}+\mathrm{0} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}\right)−\mathrm{sin}^{−\mathrm{1}} \left(−\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}−\frac{−\pi}{\mathrm{2}}\right) \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$

Commented by MJS_new last updated on 22/Oct/20

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

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