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  • Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 39 textbook solution.

Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 39 textbook solution.

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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