If int frac{dx}{cos^3 x sqrt{2 sin 2x}} = (ta | vedclass

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Question

(B) 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}}

Vedclass · Accepted Answer

$\frac{16}{5}$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$.Using $\sin 2x = 2 \sin x \cos x$,we get:$I = \int \frac{dx}{\cos^3 x \sqrt{4 \sin x \cos x}} = \frac{1}{2} \int \frac{\sec^3 x}{\sqrt{\sin x \cos x}} dx$.Divide numerator and denominator by $\cos^2 x$ inside the integral:$I = \frac{1}{2} \int \frac{\sec^4 x}{\sqrt{	an x}} dx$.Let $	an x = t$,then $\sec^2 x dx = dt$ and $\sec^2 x = 1 + t^2$.$I = \frac{1}{2} \int \frac{1+t^2}{\sqrt{t}} dt = \frac{1}{2} \int (t^{-1/2} + t^{3/2}) dt$.$I = \frac{1}{2} [2t^{1/2} + \frac{2}{5} t^{5/2}] + k = t^{1/2} + \frac{1}{5} t^{5/2} + k$.Substituting $t = 	an x$,we get $I = (	an x)^{1/2} + \frac{1}{5}(	an x)^{5/2} + k$.Comparing with $(	an x)^A + C(	an x)^B + k$,we have $A = 1/2$,$B = 5/2$,and $C = 1/5$.Thus,$A+B+C = \frac{1}{2} + \frac{5}{2} + \frac{1}{5} = 3 + \frac{1}{5} = \frac{16}{5}$.

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + k$ where $k$ is a constant of integration,then $A+B+C$ equals

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