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AP EAPCET 06-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 73 of 160

Marks: +1, -0

\int \frac{1}{\cos 4x \cos 2x} dx = \frac{1}{4\sqrt{2}} \log \left( \frac{1+f(x)}{1-f(x)} \right) - \frac{1}{2} \log g(x) + C

g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right)

$\frac{\pi}{2 \sqrt{2}}$
$\pi + 3$
$2$
$1$

Solution: 👈: Video Solution

Let

I = \int \frac{1}{\cos 4x \cos 2x} dx

I = 2 \int \frac{1}{2 \cos 4x \cos 2x} dx

= 2 \int \frac{1}{\cos 6x + \cos 2x} dx

= 2 \int \frac{1}{4 \cos^3 2x - 2 \cos 2x} dx

[\because \cos 3\theta = 4\cos^3\theta - 2\cos\theta]

= 2 \int \frac{1}{2 \cos 2x (2 \cos^2 2x - 1)} dx

= \int \frac{1}{\cos 2x (2 \cos^2 2x - 1)}

Let

\frac{1}{\cos 2x (2 \cos^2 2x - 1)} = \frac{A}{\cos 2x} + \frac{B(\cos 2x) + C}{2 \cos^2 2x - 1}

\Rightarrow 1 = (2A + B) \cos^2 2x + C \cos 2x - A

\Rightarrow A = -1, B = 2

and

C = 0

\therefore I = \int \frac{1}{\cos 2x (2 \cos^2 2x - 1)} dx

= \int \left( \frac{-1}{\cos 2x} + \frac{2 \cos 2x}{2 \cos^2 2x - 1} \right) dx

\therefore I = \int -\sec 2x \, dx + \int \frac{2 \cos 2x}{1 - 2 \sin^2 2x} dx

I = \frac{-1}{2} \log |\sec 2x + \tan 2x| + I_1

Now,

I_1 = \int \frac{2 \cos 2x \, dx}{1 - 2 \sin^2 2x}

Let

\sin 2x = t

\Rightarrow 2 \cos 2x \, dx = dt

\therefore I_1 = \int \frac{dt}{1 - 2t^2} = \frac{1}{2} \int \frac{dt}{\left( \frac{1}{\sqrt{2}} \right)^2 - t^2}

\Rightarrow I_1 = \frac{1}{2} \cdot \frac{1}{2 \cdot \left( \frac{1}{\sqrt{2}} \right)} \log \left| \frac{\frac{1}{\sqrt{2}} + t}{\frac{1}{\sqrt{2}} - t} \right| + C

I_1 = \frac{1}{2 \sqrt{2}} \log \left| \frac{1 + \sqrt{2} t}{1 - \sqrt{2} t} \right| + C

I_1 = \frac{1}{2 \sqrt{2}} \log \left| \frac{1 + \sqrt{2} \sin 2x}{1 - \sqrt{2} \sin 2x} \right| + C \quad [\because t = \sin 2x]

\Rightarrow I = \frac{1}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2} \sin 2x}{1-\sqrt{2} \sin 2x} \right| - \frac{1}{2} \log

\left| \sec 2x + \tan 2x \right| + C

\Rightarrow f(x) = \sqrt{2} \sin 2x

and

g(x) = \left| \sec 2x + \tan 2x \right|

Now,

f\left(\frac{\pi}{6}\right) = \sqrt{2} \sin \frac{\pi}{3} = \sqrt{2} \cdot \left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{\sqrt{2}}

and

g\left(\frac{\pi}{6}\right) = \left| \sec \frac{\pi}{3} + \tan \frac{\pi}{3} \right|

= |2 + \sqrt{3}| = 2 + \sqrt{3}

\therefore g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right)

= 2 + \sqrt{3} - \sqrt{2} \left( \frac{\sqrt{3}}{\sqrt{2}} \right)

= 2 + \sqrt{3} - \sqrt{3}

= 2

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