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Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 1 maths textbook solution.

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Answer : I=\frac{\pi}{4}

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

Given : \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} d x

Solution : \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} d x= I

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan \left(\frac{\pi}{2}-x\right)} d x \\ &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x \\ &I=\int_{0}^{\frac{\pi}{2}} \frac{\tan x}{1+\tan x} d x \end{aligned}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan x-1}{1+\tan x} d x \\ &I=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan x}{1+\tan x} d x-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} d x \\ &I=\int_{0}^{\frac{\pi}{2}} 1 . d x-I \end{aligned}

\begin{aligned} &I+I=[x]_{0}^{\pi / 2} \\ &\qquad \begin{array}{l} 2 I=\left[\frac{\pi}{2}-0\right] \\ I=\frac{\pi}{4} \end{array} \end{aligned}

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