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  • Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 14

Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 14

Answers (1)

Answer:
The correct answer is -e^{\frac{-x}{2}} \sec \left(\frac{x}{2}\right)+c
Hint:

I=\int e^{x}[f(x)+{f}'(x)]dx=e^{x}f(x)+c

Given:

\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{\frac{-x}{2}} d x

Solution:

        \begin{aligned} &I=\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{\frac{-x}{2}} d x \\ &\operatorname{Let}\left(\frac{-x}{2}\right)=t \quad x=-2 t \\ &\therefore d x=-2 d t \end{aligned}

        I=\int e^{t} \frac{\sqrt{(1-\sin (-2 t)}}{(1+\cos (-2 t)}(-2dt)

            =\int e^{t} \frac{\sqrt{(} 1+\sin (2 t)}{(1+\cos (2 t)}(-2 d t)

            =\int e^{t} \frac{(\cos t+\sin t)}{\cos ^{2} t}(-2 d t)

            =(-1) \int e^{t}\left\{f(t)+f^{\prime}(t)\right\} d t

Whose  f(t)=\sec t  and solution is given by e^{t} f(t)+c

        \begin{aligned} &\therefore I=2e^{t} \operatorname{sect}(-1)+c \\ &\therefore I=-2e^{t} \sec t+c \end{aligned}

So, the correct answer is -2e^{\frac{-x}{2}} \sec \left(\frac{x}{2}\right)+c

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