Question

The value of definite integral is ${\int }_0^{{\frac{\pi }{4}}^2}\frac{dx}{1+\sin \sqrt{x}+\cos \sqrt{x}}$

• A. $\pi \ln 2$
• B. $\frac{\pi \ln 2}{2}$
• C. $\frac{\pi \ln 2}{4}$
• D. $2\pi \ln 2$

Answer 1

The correct option is B $\frac{\pi \ln 2}{2}$

$I={\int }_0^{\frac{{\pi }^4}{4}}\frac{dx}{1+\sin \sqrt{x}+\cos \sqrt{x}}$

Let, $\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2t.dt$

$={\int }_0^{\pi /2}\frac{2t.dt}{1+\sin t+\cos t}={\int }_0^{\pi /2}\frac{2(\pi /2-t)dt}{1+\sin (\frac{\pi }{2}-t)+\cos (\frac{\pi }{2}-t)}$

$=2{\int }_o^{\pi /2}\frac{(\pi /2-t)dt}{1+\sin t+\cos t}=\pi .{\int }_0^{\pi /2}\frac{dt}{1+\sin t+\cos t}-2{\int }_0^{\frac{\pi }{2}}\frac{t.dt}{1+\sin t+\cos t}=\pi .{\int }_0^{\pi /2}\frac{dt}{(1+\cos t)+\sin t}-I$

$\Rightarrow I+I=\pi .{\int }_0^{\pi /2}\frac{dt}{2{\cos }^{2}t/2+2{\sin }^{2}t/2.\cos t/2}\Rightarrow 2I=\pi .{\int }_0^{\pi /2}\frac{dt}{2{\cos }^{2}t/2(1+\tan t/2)}\Rightarrow 2I=\pi .{\int }_0^{\pi /2}\frac{\frac{1}{2}{\sec }^{2}t/2.dt}{(1+\tan t/2)}\Rightarrow 2I=\pi .{\int }_0^{\pi /2}\frac{d(1+\tan t/2)}{(1+\tan t/2)}\Rightarrow 2I=\pi .[\ln |1+\tan t/2|{]}_0^{\pi /2}\Rightarrow 2I=\pi .[\ln (1+1)-0]$

$\Rightarrow 2I=\pi \ln 2\Rightarrow I=\frac{\pi }{2}\ln 2$