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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 69 Maths Textbook Solution.

$\frac{2}{3} \tan ^{-1} \frac{\operatorname{5tan}\left(\frac{x}{2}\right)-4}{3}+c$

Hint:

To solve the given statement write $\sin ^{2}\left(\frac{x}{2}\right)+\cos ^{2}\left(\frac{x}{2}\right) \text { and } \sin \operatorname{as} 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$

Given:

$\int \frac{1}{5-4 \sin x} d x$

Solution:

$I=\int \frac{1}{5-4 \sin x} d x$

$=\int \frac{1}{5\left(\sin ^{2}\left(\frac{x}{2}\right)+\cos ^{2}\left(\frac{x}{2}\right)-\frac{4}{5}\left(2 \sin \frac{x}{2} \cos \left(\frac{x}{2}\right)\right)\right.} d x$

$=\int \frac{d x}{5 \sin ^{2}\left(\frac{x}{2}\right)+5 \cos ^{2}\left(\frac{x}{2}\right)-8 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}$

$\text { Divide numerator and denominator by } \cos ^{2}\left(\frac{x}{2}\right)$

$=\int \frac{\sec ^{2}\left(\frac{x}{2}\right)}{\operatorname{stan}^{2}\left(\frac{x}{2}\right)+5-8 \tan \left(\frac{x}{2}\right)} d x$

$\text { let } \tan \left(\frac{x}{2}\right)=t$

$\frac{\sec ^{2}\left(\frac{x}{2}\right)}{2} d x=d t$

$=2 \int \frac{1}{5 t^{2}+5-8 t} d t$

$=2 \times \frac{1}{5} \int \frac{1}{t^{2} \frac{(-8 t)}{5}+1} d t$

$=\frac{2}{5} \int \frac{1}{t^{2}-\frac{8 t}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} d t$

$=\frac{2}{5} \int \frac{d t}{\left(t-\left(\frac{4}{5}\right)^{2}\right)+\left(\frac{3}{5}\right)^{2}}$

$=\frac{2}{5} \times \frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{t-\left(\frac{4}{5}\right)}{\frac{3}{5}}\right)+c$

$=\frac{2}{3} \tan ^{-1} \frac{(5 t-4)}{3}+c$

$=\frac{2}{3} \tan ^{-1} \frac{\operatorname{5tan}\left(\frac{x}{2}\right)-4}{3}+c$