Question

Solve $\int \frac{{\cos }^2\theta d\theta }{{\cos }^2\theta +4{\sin }^2\theta }$.

Answer 1

Solution :-

$I=\int \frac{co{s}^{2}x}{co{s}^{2}x+4si{n}^{2}x}dx$

$=\int \frac{co{s}^{2}x}{co{s}^{2}x+4si{n}^{2}x}.\frac{se{c}^{4}x}{se{c}^{4}x}dx$

$=\int \frac{se{c}^{2}xdx}{se{c}^{2}x+4ta{n}^{2}xse{c}^{2}x}$

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

Now $tanx=t$

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

$\therefore I=\int \frac{dt}{(1+t^2)(1+4t^2)}=\int (\frac{A}{1+t^2}+\frac{B}{1+4t^2})dt$

Now comparing $1=A(1+4t^2)+B(1+t^2)$

$\Rightarrow A+B=1$ $4A+B=0$

$\therefore A=-1/3$ $B=4/3$

$I=\frac{-1}{3}\int \frac{1}{1+t^2}dt+\frac{4}{3}\int \frac{1}{1+4t^2}dt$

We know, $\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$

$I=\frac{-1}{3}ta{n}^{-1}t+\frac{4}{3}\frac{ta{n}^{-1}\left(2t\right)}{2}+c$

Putting the value of $t$ as $\tan x$

$I=\frac{-1}{3}ta{n}^{-1}\left(tanx\right)+\frac{2}{3}ta{n}^{-1}\left(2tanx\right)+c$

$=\frac{-1}{3}x+\frac{2}{3}ta{n}^{-1}\left(2tanx\right)+c$