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  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 12 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 12 maths

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Answer:\frac{2}{\sqrt{\cos x}}+c

Hint: Use substitution method to solve this integral.

Given: \int \frac{\tan x}{\sqrt{\cos x}} d x

Solution:

        \begin{aligned} &\text { Let } I=\int \frac{\tan x}{\sqrt{\cos x}} d x \\ &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ &\Rightarrow d x=\frac{-d t}{\sin x} \text { then } \end{aligned}

        \begin{aligned} I &=\int \frac{\sin x}{t \sqrt{t}} \frac{-d t}{\sin x} \\ &=-\int \frac{1}{t^{1+\frac{1}{2}}} d t=-\int \frac{1}{t^{\frac{3}{2}}} d t \\ I &=-\int t^{\frac{-3}{2}} d t \end{aligned}

            \begin{aligned} &=-\frac{t^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+c=-\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=-(-2) \frac{1}{\sqrt{t}}+c=\frac{2}{\sqrt{t}}+c \\ &=\frac{2}{\sqrt{\cos x}}+c\; \; \; \; \; \; [\because t=\cos x] \end{aligned}

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