$\int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$ is equal to

• A

  $50 \pi$
• B

  $25 \pi$
• Correct answer
  C

  $75 \pi$
• D

  $150 \pi$

Solution

Let $\begin{aligned} I & =\int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x \\ & =100 \int_0^{\pi / 2} \frac{(\sin x+\cos x)+\sin x}{\sin x+\cos x} d x \\ & =100\left[\int_0^{\pi / 2} 1 d x+\int_0^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x\right]\end{aligned}$
$=\int_0^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x$
On adding Eqs. (i) and (ii), we get $\begin{aligned} & 2 I_1=\int_0^{\pi / 2} 1 d x=\frac{\pi}{2} \\ & \Rightarrow \quad I_1=\frac{\pi}{4} \\ & \therefore \quad I=100\left[\frac{\pi}{2}+\frac{\pi}{4}\right]=100 \times \frac{3 \pi}{4}=75 \pi \\ & \end{aligned}$

Hence, the correct answer is option: 'C'.