Mathematics

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


SOLUTION
Solution :-
$$\displaystyle I=\int \frac{cos^{2}x}{cos^{2}x+4sin^{2}x}dx$$

$$\displaystyle =\int \frac{cos^{2}x}{cos^{2}x+4sin^{2}x}.\frac{sec^{4}x}{sec^{4}x}dx$$

$$\displaystyle =\int \frac{sec^{2}x\:dx}{sec^{2}x+4tan^{2}xsec^{2}x}$$

$$\displaystyle =\int \frac{sec^{2}x\:dx}{sec^{2}x(1+4tan^{2}x)}$$

Now $$tan x=t$$

$$sec^{2}xdx=dt$$

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

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

$$\Rightarrow A+B=1$$   $$4A+B=0$$

$$\therefore A= -1/3$$    $$B= 4/3$$

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

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

$$I\displaystyle =\frac{-1}{3}tan^{-1}t+\frac{4}{3}\frac{tan^{-1}(2t)}{2}+c$$

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

$$I\displaystyle =\frac{-1}{3}tan^{-1}(tanx)+\frac{2}{3}tan^{-1}(2tanx)+c$$

$$\displaystyle =\frac{-1}{3}x+\frac{2}{3}tan^{-1}(2tanx)+c$$
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Subjective Medium Published on 17th 09, 2020
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