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  • Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 31

Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 31

Answers (1)

Answer:

\frac{\pi}{4}

Hint:

To solve this equation we should simplify cot x.

Given:

\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x

Solution:

Let

\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x

I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\cos x}{\sin x}} d x, \quad\left[\therefore \cot x=\frac{\cos x}{\sin x}\right]

=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\cos x+\sin x} d x \ldots(i), \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]

=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{2}-x\right)+\sin \left(\frac{\pi}{2}-x\right)} d x

=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i i)

Adding (i) and (ii)

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

\begin{aligned} &I=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}

 

 

Posted by

infoexpert26

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