int {frac{{dx}}{{{{cos }^3}xsqrt {2sin 2x} }} | vedclass

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(A) Let $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $.Using $\sin 2x = 2\sin x \cos x$,we get $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {

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$\sqrt {	an x} + \frac{1}{5}{	an ^{5/2}}x + c$

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(A) Let $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $.Using $\sin 2x = 2\sin x \cos x$,we get $I = \int {\frac{{dx}}{{{{\cos }^3}x\sqrt {4\sin x \cos x} }}} = \frac{1}{2}\int {\frac{{dx}}{{{{\cos }^{7/2}}x{{\sin }^{1/2}}x}}} $.This can be written as $I = \frac{1}{2}\int {\frac{{{{\sec }^4}x}}{{\sqrt {	an x} }}dx} = \frac{1}{2}\int {\frac{{(1 + {{	an }^2}x){{\sec }^2}x}}{{\sqrt {	an x} }}dx} $.Substitute $	an x = t$,so ${\sec ^2}x dx = dt$.$I = \frac{1}{2}\int {\frac{{1 + {t^2}}}{{\sqrt t }}dt} = \frac{1}{2}\int {({t^{ - 1/2}} + {t^{3/2}})dt} $.$I = \frac{1}{2} [2{t^{1/2}} + \frac{2}{5}{t^{5/2}}] + c = {t^{1/2}} + \frac{1}{5}{t^{5/2}} + c$.Substituting back $t = 	an x$,we get $I = \sqrt {	an x} + \frac{1}{5}{	an ^{5/2}}x + c$.

$\int {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} $ is equal to

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