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 16 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 16 Maths Textbook Solution.

Answers (1)

Answer: 2

Hint: Use indefinite formula then put the limit to get the required answer

Given: \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin x}dx

Solution:

\int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin x}dx

\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}\right) d x \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1-\sin x}{1-\sin ^{2} x}\right) d x \end{aligned}\left [ \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2}\right ]

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

\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right) d x \text {. } \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sec ^{2} x-\frac{\sin x}{\cos x} \frac{1}{\cos x}\right) d x \end{aligned} \quad\left[\begin{array}{l} \frac{1}{\cos x}=\sec x \\ \frac{\sin x}{\cos x}=\tan x \end{array}\right]

\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sec ^{2} x-\tan x \sec x\right) d x \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec x \tan x d x \end{aligned}                                                                    \left[\begin{array}{l} \int \sec x \tan x d x=\sec x \\ \int \sec ^{2} x d x=\tan x \end{array}\right]

=[\tan x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}-[\sec x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}

\begin{aligned} &=\left[\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)\right]-\left[\sec \frac{\pi}{4}-\sec \left(\frac{-\pi}{4}\right)\right] \\ &=\left[\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}\right)\right]-\left[\sec \frac{\pi}{4}-\sec \left(\frac{\pi}{4}\right)\right] \end{aligned} \quad\left[\begin{array}{l} \tan (-\theta)=-\tan \theta] \\ \sec (-\theta)=\sec \theta \end{array}\right]

=\left[\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{4}=1\right]

=2\tan \frac{\pi }{4}

=2.1

=2

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 Datesheet 2027 LIVE: Class 10, 12 board exam dates soon; tentative time table, sample papers
anam-khan

CBSE Answer Sheets: CIC tells board to charge photocopying fee as per RTI rules, amend circular
anam-khan

CBSE Class 12 Supplementary Result 2026: Scanned copy, verification, re-evaluation dates announced

Latest Question