# Get Answers to all your Questions

### Answers (1)

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

Hints:-  You must know the integral rules of trignometric functions.

Given:-  $\int_{0}^{\pi / 2} \frac{a \sin x+b \cos x}{\sin x+\cos x} d x$

Solution : $I=\int_{0}^{\pi / 2} \frac{a \cdot \sin x+b \cos x}{\sin x+\cos x}$                                         ....(1)

$I=\int_{0}^{\pi / 2} \frac{a \cdot \sin (\pi / 2-x)+b \cos \left(\pi / 2^{-x)}\right.}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x$

$I=\int_{0}^{\pi / 2} \frac{a \cdot \cos x+b \sin x}{\sin x+\cos x} d x$                                   .....(2)

Adding both

\begin{aligned} &2 I=\int_{0}^{\pi / 2}(a+b) \frac{\cos x+\sin x}{\sin x+\cos x} d x \\ &2 I=\int_{0}^{\pi / 2}(a+b) \cdot d x \\ &2 I=(a+b)[x]_{0}^{\pi / 2} \end{aligned}

\begin{aligned} &2 I=(a+b)\left(\frac{\pi}{2}\right) \\ &I=\frac{(a+b) \pi}{4} \end{aligned}

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