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

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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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