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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Multiple Choice Questions Question 30 Maths Textbook Solution.

Answers (1)

Answer:

x-\tan x+C

Given:

\int \frac{\cos 2 x-1}{\cos 2 x+1} d x

Hint:

Using \int \tan ^{2} x d x

Explanation:

Let \mathrm{I}=\int \frac{\cos 2 x-1}{\cos 2 x+1} d x

        =\int \frac{\cos ^{2} x-\sin ^{2} x-1}{\cos ^{2} x-\sin ^{2} x+1} d x                                                        \left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right]

        \begin{aligned} &=\int \frac{\left(\cos ^{2} x-1\right)-\sin ^{2} x}{\cos ^{2} x+\left(1-\sin ^{2} x\right)} d x \\ &=\int \frac{-\sin ^{2} x-\sin ^{2} x}{\cos ^{2} x+\cos ^{2} x} d x \end{aligned}                                                      \left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \Rightarrow 1-\sin ^{2} x=\cos ^{2} x \end{array}\right]

        \begin{aligned} &=\int \frac{-2 \sin ^{2} x}{2 \cos ^{2} x} d x \\ &=-\int \tan ^{2} x d x \end{aligned}                                                                              \left[\because \tan x=\frac{\sin x}{\cos x}\right]

         \begin{aligned} &=-\int\left(\sec ^{2} x-1\right) d x \\ &=\int\left(1-\sec ^{2} x\right) d x \\ &=\int 1 d x-\int \sec ^{2} x d x \\ &=x-\tan x+C \end{aligned}

Hence,\int \frac{\cos 2 x-1}{\cos 2 x+1} d x=x-\tan x+C

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