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#### Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 19 Maths Textbook Solution.

Answer: $\frac{1}{1-\tan (x)}+c$

Hint: $\text { To solve this we use } \tan x \text { as } t \text { and } \frac{1}{\cos x}=\sec x$

Given: $\int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} d x$

Solution: $\int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} d x$                                                $\left[\tan x=\frac{\sin x}{\cos x}\right]$

$I=\int \frac{1}{\cos ^{2} x\left(1-\frac{\sin x}{\cos x}\right)^{2}} d x$

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

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