int frac{sin alpha}{sqrt{1 + cos alpha}} d al | vedclass

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Question

(A) We are given the integral $I = \int \frac{\sin \alpha}{\sqrt{1 + \cos \alpha}} d \alpha$. Using the trigonometric identity $1 + \cos \alpha = 2 \c

Vedclass · Accepted Answer

$-2 \sqrt{2} \cos (\frac{\alpha}{2}) + c$

Vedclass · Answer

(A) We are given the integral $I = \int \frac{\sin \alpha}{\sqrt{1 + \cos \alpha}} d \alpha$. Using the trigonometric identity $1 + \cos \alpha = 2 \cos^2 (\frac{\alpha}{2})$,we substitute this into the integral: $I = \int \frac{2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})}{\sqrt{2 \cos^2 (\frac{\alpha}{2})}} d \alpha$ Assuming $\cos (\frac{\alpha}{2}) > 0$,the denominator becomes $\sqrt{2} \cos (\frac{\alpha}{2})$: $I = \int \frac{2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})}{\sqrt{2} \cos (\frac{\alpha}{2})} d \alpha$ $I = \sqrt{2} \int \sin (\frac{\alpha}{2}) d \alpha$ Integrating $\sin (\frac{\alpha}{2})$ with respect to $\alpha$: $I = \sqrt{2} 	imes (-2 \cos (\frac{\alpha}{2})) + c$ $I = -2 \sqrt{2} \cos (\frac{\alpha}{2}) + c$.

$\int \frac{\sin \alpha}{\sqrt{1 + \cos \alpha}} d \alpha =$

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