int frac{sec x}{sqrt{sin (2 x + theta) + sin | vedclass

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(B) दिया गया है $I = \int \frac{\sec x}{\sqrt{\sin (2 x + 	heta) + \sin 	heta}} d x$सूत्र $\sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C

Vedclass · Accepted Answer

$\sqrt{2 (	an x + 	an 	heta) \sec 	heta} + c$

Vedclass · Answer

(B) दिया गया है $I = \int \frac{\sec x}{\sqrt{\sin (2 x + 	heta) + \sin 	heta}} d x$सूत्र $\sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2})$ का उपयोग करने पर:$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \sin (\frac{2 x + 2 	heta}{2}) \cos (\frac{2 x + 	heta - 	heta}{2})}} d x$$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \sin (x + 	heta) \cos x}} d x$$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 (\sin x \cos 	heta + \cos x \sin 	heta) \cos x}} d x$वर्गमूल के अंदर अंश और हर को $\cos x$ से विभाजित करने पर:$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \cos^2 x (	an x \cos 	heta + \sin 	heta)}} d x$$\Rightarrow I = \int \frac{\sec x}{\sqrt{2} \cos x \sqrt{	an x \cos 	heta + \sin 	heta}} d x$$\Rightarrow I = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x}{\sqrt{	an x \cos 	heta + \sin 	heta}} d x$माना $t = 	an x \cos 	heta + \sin 	heta$,तब $dt = \sec^2 x \cos 	heta d x$,अतः $\sec^2 x d x = \frac{dt}{\cos 	heta}$.$\Rightarrow I = \frac{1}{\sqrt{2} \cos 	heta} \int t^{-1/2} dt$$\Rightarrow I = \frac{1}{\sqrt{2} \cos 	heta} \cdot 2 t^{1/2} + C$$\Rightarrow I = \sqrt{\frac{2}{\cos^2 	heta}} \sqrt{	an x \cos 	heta + \sin 	heta} + C$$\Rightarrow I = \sqrt{2 \sec^2 	heta (	an x \cos 	heta + \sin 	heta)} + C$$\Rightarrow I = \sqrt{2 \sec 	heta (	an x + 	an 	heta)} + C$

$\int \frac{\sec x}{\sqrt{\sin (2 x + \theta) + \sin \theta}} d x =$

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