Question Number 212051 by Spillover last updated on 28/Sep/24
Answered by Spillover last updated on 28/Sep/24
Answered by Ghisom last updated on 28/Sep/24
![∫_0 ^(π/2) (√(1+sin x)) dx= [t=tan (x/2) +sec (x/2)] =4∫_1 ^(1+(√2)) ((t^2 +2t−1)/((t^2 +1)^2 ))dt= =−4[((t+1)/(t^2 +1))]_1 ^(1+(√2)) =2](Q212069.png)
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:+\mathrm{sec}\:\frac{{x}}{\mathrm{2}}\right] \\ $$$$=\mathrm{4}\underset{\mathrm{1}} {\overset{\mathrm{1}+\sqrt{\mathrm{2}}} {\int}}\:\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$=−\mathrm{4}\left[\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right]_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} =\mathrm{2} \\ $$
Commented by Spillover last updated on 28/Sep/24
$${great}.{thanks} \\ $$
Answered by BaliramKumar last updated on 29/Sep/24
![∫_0 ^(π/2) (√(1 + sinx)) dx = ∫_0 ^(π/2) (√(1 + cos((π/2) − x))) dx ∫_0 ^(π/2) (√(2cos^2 ((π/4) − (x/2)))) dx (√2)∫_0 ^(π/2) cos((π/4) − (x/2)) dx −2(√2)[sin((π/4) − (x/2))]_0 ^(π/2) = 2](Q212092.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}\:+\:{sinx}}\:{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}\:+\:{cos}\left(\frac{\pi}{\mathrm{2}}\:−\:{x}\right)}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\:−\:\frac{{x}}{\mathrm{2}}\right)}\:{dx} \\ $$$$\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\frac{\pi}{\mathrm{4}}\:−\:\frac{{x}}{\mathrm{2}}\right)\:{dx} \\ $$$$−\mathrm{2}\sqrt{\mathrm{2}}\left[{sin}\left(\frac{\pi}{\mathrm{4}}\:−\:\frac{{x}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\:\mathrm{2} \\ $$