Evaluate the following intregals: 3x dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{(4\cos^3\text{x}-3\cos\text{x})}\ \text{dx}$ $\big[\cos3\text{A}=4\cos^3\text{A}-3\cos\text{A}\big]$
$=\int\frac{1}{4\cos^2\text{x}-3}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\Rightarrow\text{I}=\int\frac{\sec^2\text{x}}{4-3\sec^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4-3(1+\tan^2\text{x})}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1-3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1-(\sqrt{3}\tan\text{x})^2}\ \text{dx}$
Let $\sqrt{3}\tan\text{x}=\text{t}$
$\Rightarrow\sqrt{3}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x}\text{ dx}=\frac{\text{dt}}{\sqrt{3}}$
$\therefore\text{I}=\frac{1}{\sqrt{3}}\int\frac{\text{dt}}{1^2-\text{t}^2}$
$=\frac{1}{\sqrt{3}}\times\frac{1}{2}\ln\big|\frac{1+\text{t}}{1-\text{t}}\big|+\text{C}$
$=\frac{1}{2\sqrt{3}}\ln\Big|\frac{1+\sqrt{3}\tan\text{x}}{1-\sqrt{3}\tan\text{x}}\Big|+\text{C}$

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