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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 19

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Answer: \frac{2}{5}(\tan x)^{\frac{5}{2}}+c

Hint: Use substitution method to solve this integral.

Given:\int \tan ^{\frac{3}{2}} x \sec ^{2} x \;d x


        \begin{aligned} &\text { Let } I=\int \tan ^{\frac{3}{2}} x \sec ^{2} x \;d x \\ &\text { Put } \tan x=t \Rightarrow \sec ^{2} x\; d x=d t \text { then } \end{aligned}

        \begin{aligned} I &=\int t^{\frac{3}{2}} d t \\ &=\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\mathrm{c} \quad \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}

            \begin{aligned} &=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+c=\frac{2}{5} t^{\frac{5}{2}}+c \\ &=\frac{2}{5}(\tan x)^{\frac{5}{2}}+c \quad[\because t=\tan x] \end{aligned}

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