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

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Answer: \frac{1}{2} \tan (2 x-3)-x+c

HInt: \text { To solve this we convert tan to sec form }

Given: \int \tan ^{2}(2 x-3) d x

Solution: \int \tan ^{2}(2 x-3) d x

\begin{aligned} &=\int \sec ^{2}(2 x-3)-1 d x\left[\because \tan ^{2} x=\sec ^{2} x-1\right] \\ &=\int \sec ^{2}(2 x-3)-\int 1 d x \\ &=\frac{\tan (2 x-3)}{2}-x+c\left[\because \int \sec ^{2}(a x+b) d x=\frac{\tan (a x+b)}{a}+c\right] \\ &=\frac{1}{2} \tan (2 x-3)-x+c \end{aligned}


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