Question

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

Answer 1

$I=\int \frac{cosx}{cos3x}dx=\int \frac{cosx}{4co{s}^{3}x-3cosx}dx=\int \frac{du}{4co{s}^{2}x-3}$

$I=\int \frac{dx}{co{s}^{2}x[4-\frac{3}{co{s}^{2}x}]}$

$=\int \frac{se{c}^{2}xdx}{4-3se{c}^{2}x}$

$I=\int \frac{se{c}^{2}xdx}{4-3(1+ta{n}^{2}x)}=\int \frac{se{c}^{2}xdx}{1-3ta{n}^{2}x}$

put $tanx=t$

$se{c}^{2}xdx=dt$

$\therefore I=\int \frac{dt}{1-3t^2}=\int \frac{dt}{(1+\sqrt{3}t)(1-\sqrt{3}t)}$

$I=\frac{1}{2}\int [\frac{1}{(1-\sqrt{3}t)}+\frac{1}{(1+\sqrt{3}t)}]dt$

$I=\frac{1}{2}\times [-\frac{1}{\sqrt{3}}ln|1-\sqrt{3}t|+\frac{1}{\sqrt{3}}ln|1+\sqrt{3}t|]+ln\left|c\right|$

$I=\frac{1}{2\sqrt{3}}ln\left|\frac{1+\sqrt{3}t}{1-\sqrt{3}t}\right|+ln\left|c\right|$

$\therefore I=\frac{1}{2\sqrt{3}}ln\left|\frac{1+\sqrt{3}tanx}{1-\sqrt{3}tanx}\right|+c$

$\int \frac{cosx}{cos3x}dx=\frac{1}{2\sqrt{3}}ln\left|\frac{1+\sqrt{3}tanx}{1-\sqrt{3}tanx}\right|+c$