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

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(A) We have the integral $I = \int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$.Using $\sin 2x = 2 \sin x \cos x$,the integral becomes:$I = \int \frac{dx}{\c

Vedclass · Accepted Answer

$\frac{16}{5}$

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(A) We have the integral $I = \int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$.Using $\sin 2x = 2 \sin x \cos x$,the integral becomes:$I = \int \frac{dx}{\cos^3 x \sqrt{4 \sin x \cos x}} = \int \frac{dx}{2 \cos^3 x \sqrt{\sin x \cos x}} = \int \frac{dx}{2 \cos^4 x \sqrt{	an x}}$.Since $\frac{1}{\cos^4 x} = \sec^4 x = (1 + 	an^2 x) \sec^2 x$,we have:$I = \frac{1}{2} \int \frac{(1 + 	an^2 x) \sec^2 x}{\sqrt{	an x}} dx$.Let $u = 	an x$,then $du = \sec^2 x dx$:$I = \frac{1}{2} \int \frac{1 + u^2}{\sqrt{u}} du = \frac{1}{2} \int (u^{-1/2} + u^{3/2}) du$.Integrating term by term:$I = \frac{1}{2} \left( \frac{u^{1/2}}{1/2} + \frac{u^{5/2}}{5/2} ight) + k = u^{1/2} + \frac{1}{5} u^{5/2} + k$.Substituting $u = 	an x$ back:$I = (	an x)^{1/2} + \frac{1}{5} (	an x)^{5/2} + k$.Comparing this with the given form $(	an x)^A + C(	an x)^B + k$,we get $A = 1/2$,$B = 5/2$,and $C = 1/5$.Therefore,$A + B + C = \frac{1}{2} + \frac{5}{2} + \frac{1}{5} = 3 + \frac{1}{5} = \frac{16}{5}$.

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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