$$ \int \frac{1}{\cos 4 x \cos 2 x} d x=\frac{1}{2 \sqrt{2}} \log \left(\frac{1+f(x)}{1-f(x)}\right) $$…
$$
\int \frac{1}{\cos 4 x \cos 2 x} d x=\frac{1}{2 \sqrt{2}} \log \left(\frac{1+f(x)}{1-f(x)}\right)
$$
$-\frac{1}{2} \log g(x)+C$, then $g\left(\frac{\pi}{6}\right)-\sqrt{2} f\left(\frac{\pi}{6}\right)=$
- $\frac{\pi}{2 \sqrt{2}}$
- $\pi+3$
- 2
- 1
Solution
$
\text { Let } \begin{aligned}
I & =\int \frac{1}{\cos 4 x \cos 2 x} d x \\
\qquad & =2 \int \frac{1}{2 \cos 4 x \cos 2 x} d x \\
& =2 \int \frac{1}{\cos 6 x+\cos 2 x} d x \\
& =2 \int \frac{1}{4 \cos ^3 2 x-2 \cos 2 x} d x
\end{aligned}
$
$\begin{aligned} & =2 \int \frac{1}{2 \cos 2 x\left(2 \cos ^2 2 x-1\right)} d x \\ & =\int \frac{1}{\cos 2 x\left(2 \cos ^2 2 x-1\right)} \\ & \text { Let } \frac{1}{\cos 2 x\left(2 \cos ^2 2 x-1\right)}=\frac{A}{\cos 2 x}+\frac{B(\cos 2 x)+C}{2 \cos ^2 2 x-1} \\ & \Rightarrow \quad 1=(2 A+B) \cos ^2 2 x+C \cos 2 x-A \\ & \Rightarrow \quad A=-1, B=2 \text { and } C=0 \\ & \therefore \quad I=\int \frac{1}{\cos 2 x\left(2 \cos ^2 2 x-1\right)} d x \\ & =\int\left(\frac{-1}{\cos 2 x}+\frac{2 \cos 2 x}{2 \cos ^2 2 x-1}\right) d x \\ & \therefore \quad I=\int-\sec 2 x d x+\int \frac{2 \cos 2 x}{1-2 \sin ^2 2 x} d x \\ & I=\frac{-1}{2} \log |\sec 2 x+\tan 2 x|+I_1 \\ & \end{aligned}$
Now,
$
I_1=\int \frac{2 \cos 2 x d x}{1-2 \sin ^2 2 x}
$
Let $\quad \sin 2 x=t$
$
\begin{aligned}
& \Rightarrow 2 \cos 2 x d x=d t \\
& \therefore \quad I_1=\int \frac{d t}{1-2 t^2}=\frac{1}{2} \int \frac{d t}{\left(\frac{1}{\sqrt{2}}\right)^2-t^2} \\
& \Rightarrow \quad 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 2 x}{1-\sqrt{2} \sin 2 x}\right|+C \quad[\because t=\sin 2 x] \\
& \Rightarrow \quad I=\frac{1}{2 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \sin 2 x}{1-\sqrt{2} \sin 2 x}\right|-\frac{1}{2} \log \\
& |\sec 2 x+\tan 2 x|+C \\
& \Rightarrow \quad f(x)=\sqrt{2} \sin 2 x \text { and } g(x)=|\sec 2 x+\tan 2 x| \\
& \text { 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}} \\
& \text { 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) \\
&
\end{aligned}
$
$\begin{aligned} & =2+\sqrt{3}-\sqrt{2}\left(\frac{\sqrt{3}}{\sqrt{2}}\right) \\ & =2+\sqrt{3}-\sqrt{3} \\ & =2\end{aligned}$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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