Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 64 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 64 Maths Textbook Solution.

Answers (1)

Answer: \frac{\pi }{32}

Hint: Use indefinite formula then put the limit to solve this integral

Given:\int_{0}^{\frac{\pi}{4}}(\tan x+\cot x)^{-2} d x

Solution:

\begin{aligned} &\int_{0}^{\frac{\pi}{4}}(\tan x+\cot x)^{-2} d x=\int_{0}^{\frac{\pi}{4}} \frac{1}{(\tan x+\cot x)^{2}} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right] \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right)^{2}} d x \end{aligned}

=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x \cos x}{\sin ^{2} x+\cos ^{2} x}\right)^{2} d x                                                    \left[\sin ^{2} x+\cos ^{2} x=1\right] \\

=\int_{0}^{\frac{\pi}{4}} \frac{(\sin x \cos x)^{2}}{1} d x \\    

=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x \cos ^{2} x d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \                                                                \left[1-\sin ^{2} \theta =\cos ^{2} \theta \right] \\

=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x\left(1-\sin ^{2} x\right) d x \\                                                                \left[\begin{array}{l} 2 \sin ^{2} \theta=1-\cos 2 \theta \\ \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2} \end{array}\right]

=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x d x-\int_{0}^{\frac{\pi}{4}} \sin ^{4} x d x

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2} d x-\int_{0}^{\frac{\pi}{4}}\left(\sin ^{2} x\right)^{2} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{2}(1-\cos 2 x) d x-\int_{0}^{\frac{\pi}{4}}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}}(1-\cos 2 x) d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}(1-\cos 2 x)^{2} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}\left(1+\cos ^{2} 2 x-2 \cos 2 x\right) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}} \cos ^{2} 2 x d x-\frac{2}{4} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x \\ &=\left(\frac{1}{2}-\frac{1}{4}\right) \int_{0}^{\frac{\pi}{4}} 1 d x-\left(\frac{1}{2}+\frac{1}{2}\right) \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}\left(\frac{1+\cos 4 x}{2}\right) d x \end{aligned}

                                                                                                                                                        \left [ 1+\cos 2\theta =2\cos ^{2}\theta \right ]

\begin{aligned} &=\left(\frac{2-1}{4}\right) \int_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\left(\frac{1}{4}-\frac{1}{8}\right)_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\left(\frac{2-1}{8}\right)[x]_{0}^{\frac{\pi}{4}}-\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{4}}-\frac{1}{8}\left[\frac{\sin 4 x}{4}\right]_{0}^{\frac{\pi}{4}} \end{aligned}

                                                                                                                                                        \left [ \because \int 1dx=x,\int \cos a\: xdx=\frac{\sin ax}{a} \right ]

=\frac{1}{8}\left(\frac{\pi}{4}-0\right)-\frac{1}{2}\left[\sin \frac{2 \pi}{4}-\sin 0\right]-\frac{1}{8}\left[\sin \frac{4 \pi}{4}-\sin 0\right]

=\frac{1}{8} \frac{\pi}{4}-\frac{1}{2}\left[\sin \frac{\pi}{2}-0\right]-\frac{1}{8}[\sin \pi-\sin 0]         

                                                                                                                                            \left [ \because \sin 0=0,\sin \pi =0 \right ]                                                   

=\frac{\pi}{32}-\frac{1}{2}[0-0]-\frac{1}{8}[0-0]

=\frac{\pi}{32}

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE asks schools to ensure error-free LOC forms for Classes 10, 12; edit window from October 13
anam-khan

‘You ruined my career’: Students hit back at CBSE over 'abrupt' removal of additional subjects
anam-khan

‘Careers are at stake’: CBSE removes additional subject option for private students without prior notice

Latest Question