Question

${\int }_0^{\pi /2}\frac{a\cos x+b\sin x}{\cos x+\sin x}dx=$

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

Answer 1

Answer: (C)

$\frac{\pi }{4}(a+b)$

$I={\int }_0^{\frac{\pi }{2}}\frac{a\cos x+b\sin x}{\cos x+\sin x}dx$ ..........eq(i)
We know that, ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
So, $I={\int }_0^{\frac{\pi }{2}}\frac{a\sin x+b\cos x}{\cos x+\sin x}dx$ ............eq(ii)

Adding eq(i) and (ii)

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{(a+b)\cos x+(a+b)\sin x}{\cos x+\sin x}dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{(a+b)(\cos x+\sin x)}{\cos x+\sin x}dx$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}(a+b)dx$

$\Rightarrow 2I=(a+b){\int }_0^{\frac{\pi }{2}}1dx$

$\Rightarrow I=\frac{(a+b)}{2}\left(\frac{\pi }{2}\right)$

$\therefore I=\frac{(a+b)\pi }{4}$