Get Answers to all your Questions

header-bg qa

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.3 Question 11 Maths Textbook Solution.

Answers (1)

Answer:2 \tan \frac{x}{2}-x+c

Hint: Using \cos \frac{x}{2}=2 \cos ^{2} \frac{x}{2}-1=1-2 \sin ^{2} \frac{x}{2}

Given: \int \frac{1-\cos x}{1+\cos x} d x

Solution: \int \frac{1-\cos x}{1+\cos x} d x

=\int \frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x

=\int \tan ^{2} \frac{x}{2} d x                                                                                    \left[\tan ^{2} x=\sec ^{2} x-1\right]

=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x

\begin{aligned} &=\int \sec ^{2} \frac{x}{2} d x-\int d x \quad\left[\int \sec ^{2}(a x+b) d x=\frac{\tan (a x+b)}{a}+c\right] \\ &=\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x+c \\ &=2 \tan \frac{x}{2}-x+c \end{aligned}

 

Posted by

infoexpert21

View full answer