int sqrt{1 + sin left( frac{x}{4} right)} , d | vedclass

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Question

(A) We need to evaluate the integral $I = \int \sqrt{1 + \sin \left( \frac{x}{4} ight)} \, dx$. Using the trigonometric identities $1 = \sin^2 	het

Vedclass · Accepted Answer

$8\left( \sin \frac{x}{8} - \cos \frac{x}{8} \right) + c$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int \sqrt{1 + \sin \left( \frac{x}{4} ight)} \, dx$. Using the trigonometric identities $1 = \sin^2 	heta + \cos^2 	heta$ and $\sin(2	heta) = 2\sin 	heta \cos 	heta$,where $	heta = \frac{x}{8}$,we have: $1 + \sin \left( \frac{x}{4} ight) = \sin^2 \frac{x}{8} + \cos^2 \frac{x}{8} + 2\sin \frac{x}{8} \cos \frac{x}{8} = \left( \sin \frac{x}{8} + \cos \frac{x}{8} ight)^2$. Substituting this into the integral: $I = \int \sqrt{\left( \sin \frac{x}{8} + \cos \frac{x}{8} ight)^2} \, dx = \int \left( \sin \frac{x}{8} + \cos \frac{x}{8} ight) \, dx$. Integrating term by term: $I = \frac{-\cos(x/8)}{1/8} + \frac{\sin(x/8)}{1/8} + c = 8\left( \sin \frac{x}{8} - \cos \frac{x}{8} ight) + c$.

$\int \sqrt{1 + \sin \left( \frac{x}{4} \right)} \, dx$

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