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

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

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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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