Mathematics

Solve $$\displaystyle\int { \dfrac { \sin { x }  }{ 1-4\cos ^{ 2 }{ x }  }  } .dx.$$


SOLUTION
 $$\displaystyle\int { \dfrac { \sin { x }  }{ 1-4\cos ^{ 2 }{ x }  }  } .dx.$$

Put, $$ 2\cos x = t $$

$$\implies - 2 \sin x dx = dt $$

Now, $$\displaystyle\int \dfrac{\sin x}{1 - 4\cos^2 x} dx = \dfrac{1}{2} \int \dfrac{-1}{1 - t^2} dt$$

$$ = \dfrac{1}{2} \displaystyle\int \dfrac{1}{t^2 - 1} dt$$

$$ = \dfrac{1}{2}\times \dfrac{1}{2} \log \left|{\dfrac{t - 1}{t + 1}} \right|+ c$$

$$ =  \dfrac{1}{4} \log \left|{\dfrac{2\cos x - 1}{2\cos x + 1}} \right|+ c$$ (Ans)
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Subjective Medium Published on 17th 09, 2020
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