Question Number 163158 by mnjuly1970 last updated on 04/Jan/22
$$ \\ $$$$\:\:\:\:\:{prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\:{sin}\left({x}\right)+{cos}\left({x}\right)}{\:\sqrt{\mathrm{1}+{sin}\left({x}\right){cos}\left({x}\right)}}\:{dx}=\:\sqrt{\mathrm{2}}\:.{cot}^{\:−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\right) \\ $$$$\:\:\:−−−−− \\ $$
Answered by mahdipoor last updated on 04/Jan/22
![get sinx−cosx=(√3)u { (((√3)du=(cosx+sinx)dx)),((3u^2 =sin^2 x+cos^2 x−2sinx.cosx⇒((1−3u^2 )/2)=sinx.cosx)) :} ⇒∫_0 ^( π/4) ((sinx+cosx)/( (√(1+sinx.cosx))))dx=∫_(−1/(√3)) ^( 0) (((√3)du)/( (√(1+((1−3u^2 )/2)))))= (√2)∫_( −1/(√3)) ^( 0) (du/( (√(1−u^2 ))))=[(√2)sin^(−1) u+C]_(−1/(√3)) ^0 = −(√2)sin^(−1) (((−1)/( (√3))))=(√2)sin^(−1) ((1/( (√3)))) if sinx=(1/( (√3))) ⇒ cotx=(√2) ⇒(√2)sin^(−1) ((1/( (√3))))=(√2)cot^(−1) ((√2))](https://www.tinkutara.com/question/Q163162.png)
$${get}\:{sinx}−{cosx}=\sqrt{\mathrm{3}}{u}\:\:\:\:\:\: \\ $$$$\begin{cases}{\sqrt{\mathrm{3}}{du}=\left({cosx}+{sinx}\right){dx}}\\{\mathrm{3}{u}^{\mathrm{2}} ={sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}−\mathrm{2}{sinx}.{cosx}\Rightarrow\frac{\mathrm{1}−\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}}={sinx}.{cosx}}\end{cases} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \frac{{sinx}+{cosx}}{\:\sqrt{\mathrm{1}+{sinx}.{cosx}}}{dx}=\int_{−\mathrm{1}/\sqrt{\mathrm{3}}} ^{\:\mathrm{0}} \frac{\sqrt{\mathrm{3}}{du}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}−\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}}}}= \\ $$$$\sqrt{\mathrm{2}}\int_{\:−\mathrm{1}/\sqrt{\mathrm{3}}} ^{\:\mathrm{0}} \frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}=\left[\sqrt{\mathrm{2}}{sin}^{−\mathrm{1}} {u}+{C}\right]_{−\mathrm{1}/\sqrt{\mathrm{3}}} ^{\mathrm{0}} \: \\ $$$$=\:−\sqrt{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\sqrt{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${if}\:{sinx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:{cotx}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\sqrt{\mathrm{2}}{cot}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right) \\ $$
Commented by mnjuly1970 last updated on 04/Jan/22
$${bravo}\:{sir}\:{mahdipoor} \\ $$