If int frac{sin left(x-frac{pi}{4}right)}{2+s | vedclass

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Question

(B) Given the integral $I = \int \frac{\sin \left(x-\frac{\pi}{4}\right)}{2+\sin 2 x} d x$. Using the identity $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\si

Vedclass · Accepted Answer

$\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)$

Vedclass · Answer

(B) Given the integral $I = \int \frac{\sin \left(x-\frac{\pi}{4}ight)}{2+\sin 2 x} d x$. Using the identity $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,the denominator becomes $2 + \sin 2x = 3 - (\sin x - \cos x)^2$. This is not the most direct path. Alternatively,note that $2 + \sin 2x = 1 + (1 + \sin 2x) = 1 + (\sin x + \cos x)^2$. Thus,$I = \int \frac{\sin x \cos(\pi/4) - \cos x \sin(\pi/4)}{1 + (\sin x + \cos x)^2} dx = \frac{1}{\sqrt{2}} \int \frac{\sin x - \cos x}{1 + (\sin x + \cos x)^2} dx$. Let $t = \sin x + \cos x$,then $dt = (\cos x - \sin x) dx$,which implies $-dt = (\sin x - \cos x) dx$. Substituting these into the integral: $I = \frac{1}{\sqrt{2}} \int \frac{-dt}{1 + t^2} = -\frac{1}{\sqrt{2}} 	an^{-1}(t) + C$. Substituting $t = \sin x + \cos x$ back,we get $I = -\frac{1}{\sqrt{2}} 	an^{-1}(\sin x + \cos x) + C$. Comparing this with the given form $-\frac{1}{\sqrt{2}} 	an^{-1}(f(x)) + C$,we find $f(x) = \sin x + \cos x$. Since $\sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x ight) = \sqrt{2} \cos(x - \frac{\pi}{4})$,the correct option is $B$.

If $\int \frac{\sin \left(x-\frac{\pi}{4}\right)}{2+\sin 2 x} d x=-\frac{1}{\sqrt{2}} \tan ^{-1}(f(x))+C$,then $f(x)=$

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