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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 41 maths textbook solution

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

Hint: Use substitution method to solve this integral.

Given:   \int \tan x \cdot \sec ^{2} x \sqrt{1-\tan ^{2} x}\; d x


        \begin{aligned} &\text { Let } I=\int \tan x \cdot \sec ^{2} x \sqrt{1-\tan ^{2} x} d x \\ &\text { Put } \quad 1-\tan ^{2} x=t \Rightarrow-2 \tan x \cdot \sec ^{2} x d x=d t \\ &\Rightarrow d x=\frac{1}{-2 \tan x \cdot \sec ^{2} x} \text { dt then } \end{aligned}

        \Rightarrow I=\int \tan x \cdot \sec ^{2} x \cdot \sqrt{t} \frac{1}{-2 \tan x \cdot \sec ^{2} x} \mathrm{dt}

        =-\frac{1}{2} \int \sqrt{t} d t=-\frac{1}{2} \int t^{\frac{1}{2}} d t

        =-\frac{1}{2}\left[\frac{t_{}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

        =-\frac{1}{2}\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]+c=-\frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}}+c        

        =-\frac{1}{3}\left(1-\tan ^{2} x\right)^{\frac{3}{2}}+c \quad\left[\because t=1-\tan ^{2} x\right]

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