Question

integral of cosx/cos3x

Answer 1

Hey There,
Here we are going to use the following concept:

$\cos 3x = 4\cos^3x-3\cos x \\ \\ \implies \cos 3x = \cos x(4\cos^2x-3) \\ \\ \implies \frac{\cos x}{\cos 3x} = \frac{1}{4\cos^2x-3}$

Now, our question reduces to:
$\int \frac{\cos x}{\cos 3x}dx \\ \\ \\ = \int \frac{1}{4\cos^2x-3}dx \\ \\ \\ = \int \frac{1}{\frac{4}{\sec^2x}-3}dx \\ \\ \\ = \int \frac{\sec^2x}{4-3\sec^2x}dx \\ \\ \\ = \int \frac{\sec^2x}{4-3(1+\tan^2x)}dx \\ \\ \\ = \int \frac{\sec^2x}{4-3-3\tan^2x}dx \\ \\ \\ = \int \frac{\sec^2x}{1-3\tan^2x}dx \\ \\ \\ \\ Let \, \, \tan x = t \\ \\ \implies \sec^2x dx = dt \\ \\ \\ \\ Now \\ \\ \int \frac{\sec^2x}{1-3\tan^2x}dx \\ \\ \\ = \int \frac{1}{1-3t^2}dt \\ \\ \\ = \frac{1}{3} \int \frac{dt}{\frac{1}{3}-t^2} \\ \\ \\ = \frac{1}{3} \int \frac{dt}{\left(\frac{1}{\sqrt{3}}\right)^2-t^2} \\ \\ \\ \\ \textrm{Now we use the identity: } \\ \\ \\ \\ \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| +c \\ \\ \\ \\ \textrm{So we have } \\ \\ \\ \\ \frac{1}{3} \int \frac{dt}{\left(\frac{1}{\sqrt{3}}\right)^2-t^2} \\ \\ \\ = \frac{1}{3} \times \frac{1}{2\frac{1}{\sqrt{3}}} \ln \left| \frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right|+c \\ \\ \\ = \frac{1}{2\sqrt{3}} \ln \left| \frac{1+\sqrt{3}t}{1-\sqrt{3}t} \right| +c \\ \\ \\ = \frac{1}{2\sqrt{3}} \ln \left| \frac{1+\sqrt{3}\tan x}{1-\sqrt{3}\tan x} \right| + c$



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