int frac{sqrt{2} sin x}{sin left(x+frac{pi}{4 | vedclass

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(D) Let $I = \int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$. Using the substitution $t = x+\frac{\pi}{4}$,we have $x = t-\frac{\p

Vedclass · Accepted Answer

$x-\log \left|\sin \left(x+\frac{\pi}{4}\right)\right|+c$

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(D) Let $I = \int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$. Using the substitution $t = x+\frac{\pi}{4}$,we have $x = t-\frac{\pi}{4}$ and $dx = dt$. Substituting these into the integral: $I = \int \frac{\sqrt{2} \sin \left(t-\frac{\pi}{4}\right)}{\sin t} dt$. Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $I = \int \sqrt{2} \frac{\sin t \cos \frac{\pi}{4} - \cos t \sin \frac{\pi}{4}}{\sin t} dt$. Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$: $I = \int \sqrt{2} \frac{\left(\frac{1}{\sqrt{2}} \sin t - \frac{1}{\sqrt{2}} \cos t\right)}{\sin t} dt = \int \frac{\sin t - \cos t}{\sin t} dt$. $I = \int (1 - \cot t) dt = t - \ln |\sin t| + c$. Substituting back $t = x+\frac{\pi}{4}$: $I = x + \frac{\pi}{4} - \ln |\sin (x+\frac{\pi}{4})| + c$. Since $\frac{\pi}{4}$ is a constant,it can be absorbed into the constant $c$: $I = x - \log |\sin (x+\frac{\pi}{4})| + c$.

$\int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$ is equal to

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