int frac{cos x-sin x}{5+sin (2 x)} d x= | vedclass

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Question

(B) माना $I = \int \frac{\cos x-\sin x}{5+\sin 2 x} d x$. हम जानते हैं कि $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x$.

Vedclass · Accepted Answer

$\frac{1}{2} 	an ^{-1}\left[\frac{1}{2}(\sin x+\cos x)ight]+c$

Vedclass · Answer

(B) माना $I = \int \frac{\cos x-\sin x}{5+\sin 2 x} d x$. हम जानते हैं कि $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x$. अतः,$\sin 2x = (\sin x + \cos x)^2 - 1$. इस मान को समाकलन में रखने पर: $I = \int \frac{\cos x-\sin x}{5 + (\sin x + \cos x)^2 - 1} d x = \int \frac{\cos x-\sin x}{4 + (\sin x + \cos x)^2} d x$. माना $t = \sin x + \cos x$. तब $dt = (\cos x - \sin x) dx$. समाकलन में $t$ और $dt$ रखने पर: $I = \int \frac{dt}{4 + t^2} = \frac{1}{2} 	an^{-1}\left(\frac{t}{2}ight) + C$. अंत में $t = \sin x + \cos x$ रखने पर: $I = \frac{1}{2} 	an^{-1}\left(\frac{\sin x + \cos x}{2}ight) + C$.

$\int \frac{\cos x-\sin x}{5+\sin (2 x)} d x=$

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