int frac{1}{cos ^3 x sqrt{sin 2 x}} ,d x= | vedclass

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(A) माना $I = \int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x$.चूँकि $\sin 2x = 2 \sin x \cos x$,इसलिए $I = \int \frac{1}{\cos ^3 x \sqrt{2 \sin x \cos

Vedclass · Accepted Answer

$\sqrt{2}\left(\sqrt{	an x}+\frac{1}{5}(	an x)^{\frac{5}{2}}ight)+c$,जहाँ $c$ समाकलन का एक स्थिरांक है।

Vedclass · Answer

(A) माना $I = \int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x$.चूँकि $\sin 2x = 2 \sin x \cos x$,इसलिए $I = \int \frac{1}{\cos ^3 x \sqrt{2 \sin x \cos x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{1}{\cos ^3 x \sqrt{\sin x} \sqrt{\cos x}} \,d x$.इसे सरल करने पर $I = \frac{1}{\sqrt{2}} \int \frac{1}{\cos^{3.5} x \sqrt{\sin x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{\sec^{3.5} x}{\sqrt{	an x}} \,d x$.आइए समाकल्य को $\frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 \sin x \cos x}} = \frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 	an x \cos^2 x}} = \frac{\sec^4 x}{\sqrt{2 	an x}}$ के रूप में फिर से लिखें।अतः,$I = \frac{1}{\sqrt{2}} \int \frac{\sec^4 x}{\sqrt{	an x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{(1+	an^2 x) \sec^2 x}{\sqrt{	an x}} \,d x$.माना $	an x = t$,तो $\sec^2 x \,d x = dt$.$I = \frac{1}{\sqrt{2}} \int \frac{1+t^2}{\sqrt{t}} \,dt = \frac{1}{\sqrt{2}} \int (t^{-1/2} + t^{3/2}) \,dt$.$I = \frac{1}{\sqrt{2}} \left( \frac{t^{1/2}}{1/2} + \frac{t^{5/2}}{5/2} ight) + c = \frac{1}{\sqrt{2}} \left( 2\sqrt{t} + \frac{2}{5} t^{5/2} ight) + c$.$I = \sqrt{2} \sqrt{t} + \frac{\sqrt{2}}{5} t^{5/2} + c = \sqrt{2} \left( \sqrt{	an x} + \frac{1}{5} (	an x)^{5/2} ight) + c$.

$\int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x=$

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