Question

Evaluate ${\int }_0^{\pi /2}\frac{a\sin x+b\cos x}{\sin (\frac{\pi }{4}+x)}dx$

• A. $\frac{\pi (a+b)}{4\sqrt{2}}$
• B. $\frac{\pi (a-b)}{2\sqrt{2}}$
• C. $\frac{\pi (a+b)}{2\sqrt{2}}$
• D. $\frac{\pi (a-b)}{4\sqrt{2}}$

Answer 1

Answer: (C)

$\frac{\pi (a+b)}{2\sqrt{2}}$

$l={\int }_0^{\pi /2}\frac{a\sin x+b\cos x}{\sin (\frac{\pi }{4}+x)}dx$$l={\int }_0^{\pi /2}\frac{a\sin x+b\cos x}{\sin x+\cos x}dx$....(i)
$l=\sqrt{2}{\int }_0^{\pi /2}\frac{a\cos x+b\sin x}{\cos x+\sin x}$......(ii)
Adding (1) & (2), we get
$2l=\sqrt{2}(a+b){\int }_0^{\pi /2}dx\Rightarrow l=\frac{(a+b)\pi }{2\sqrt{2}}$