Mathematics

Integrate : $$\int {\dfrac{{\cos 2x}}{{1 + {{\sin }^2}x}}dx} $$


SOLUTION
$$I=\displaystyle \int \dfrac {\cos 2x}{1+\sin^2 x}dx=\displaystyle \int \dfrac {\cos 2x}{1+\dfrac {1-\cos 2x}{2}}dx$$  [Using $$\cos 2x=1-2\sin^2 x]$$
$$I=-2\displaystyle \int \dfrac {-\cos 2x}{3-\cos 2x }$$ [Adding and subtracting both sides]
$$I=-2\displaystyle \int \dfrac {(3-\cos 2x)-3}{3-\cos 2x}dx$$ [Splitting the fraction]
$$I=-2\displaystyle \int 1.dx +6 \displaystyle \int \dfrac {1}{3-\cos 2x}dx$$
$$I=-2x+6\displaystyle \int \dfrac {1}{3-\dfrac {1-\tan^2 x}{1+\tan^2 x}}dx\quad  \left [Using\ relation\ \cos 2x =\dfrac {1-\tan^2x}{1+\tan^2x}\right]$$
$$I=-2x+6\displaystyle \int \dfrac {\sec^2 x}{2+4\tan^2 x}dx \ [\sec^2x=1+\tan^2 x]$$
let $$\tan x=t$$
Differentiation both sides:-
$$\sec^2x\ dx=dt$$
Subtracting in the integral:-
$$\Rightarrow \ I=-2x+6^3 \displaystyle \int \dfrac {dt}{x(1+2t^2)}$$
$$=-2x+3 \displaystyle \int \dfrac {dt}{1+(\sqrt 2 t^2)}\quad \left [integral\ of\ from\ \dfrac {1}{1+x^2}\quad \because \displaystyle \int \dfrac {1}{1+x^2}dx=\tan^{-1}x+c \right]$$
$$=-2x+\dfrac {3}{\sqrt 2}\tan^{-1}(\sqrt 2 t)+c$$
$$\Rightarrow \ I=-2x+\dfrac {3}{\sqrt 2}\tan^{-1}(\sqrt 2 \tan x)+c$$

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Subjective Medium Published on 17th 09, 2020
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