Question

Solve $\int \frac{cosx}{cos3x}dx=$

Answer 1

Consider the given function:

$\int \frac{\cos x}{\cos 3x}dx$

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

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

$=\int \frac{1}{4{\cos }^{2}x-3({\sin }^{2}x+{\cos }^{2}x)}dx$

$=\int \frac{1}{3.{\cos }^2-{\sin }^{2}x}dx$

$=\int \frac{1}{{\sin }^{2}x(3-{\cot }^{2}x)}dx$

$=\int \frac{\frac{1}{{\sin }^{2}x}}{3-{\cot }^{2}x}dx$

$=\int \frac{\cos e{c}^{2}x}{3-{\cot }^{2}x}dx$

$put\cot x=t$

$-\cos e{c}^{2}xdx=dt$

$=\int \frac{dt}{3-t^2}$

$=\frac{1}{2}.\frac{1}{\sqrt{3}}\log x\frac{\sqrt{3}+t}{\sqrt{3}-t}+c$

$=\frac{1}{2\sqrt{3}}\log \frac{\sqrt{3}+\cot x}{\sqrt{3}-\cot x}+c$

Hence this is the answer.