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#### Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 2 maths textbook solution.

Answer : $\frac{\pi}{4}$

Hints:-  You must know the integration rules of trigonometric functions and its limits

Given : $\int_{0}^{\pi / 2} \frac{1}{1+\cot x} d x$

Solution : $\int_{0}^{\pi / 2} \frac{1}{1+\cot x} d x$

\begin{aligned} &\int_{0}^{\pi / 2} \frac{d x}{1+\frac{\cos x}{\sin x}} \\ &\int_{0}^{\pi / 2} \frac{d x \cdot \sin x}{\sin x+\cos x} \end{aligned}                                                .....(1)

We know                                                     $\left[\int_{0}^{a} f(x)=\int_{0}^{a} f(a-x) d x\right]$

\begin{aligned} &I=\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\\ &I=\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x \end{aligned}            ....(2)

\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x+\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x \\ &2 I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sin x+\cos x} d x \\ &2 I=\int_{0}^{\pi / 2} 1 \cdot d x \end{aligned}

$2 I=[x]_{0}^{\pi / 2}$

\begin{aligned} &I=\frac{1}{2} \times \frac{\pi}{2} \\ &=\frac{\pi}{4} \end{aligned}