Mathematics

Solve $$\displaystyle\int\dfrac{{\cos x - \sin x}}{{1 + \sin 2x}}dx$$


SOLUTION

Solve the given trigonometric function.

$$ =\displaystyle\int\dfrac{cosx-sinx}{1+sin2x} dx$$

$$ =\displaystyle\int\dfrac{cosx-sinx}{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x} dx$$

$$ =\displaystyle\int\dfrac{cosx-sinx}{{{\left( cosx+sinx \right)}^{2}}} dx$$

Let $$cosx+sinx=t$$

Then,

  $$ \dfrac{d}{dx}\left( cosx+sinx \right)=\dfrac{d}{dx}t $$

 $$ \Rightarrow -\sin x+\cos x=\dfrac{dt}{dx} $$

 $$ \Rightarrow \left( \cos x-\sin x \right)dx=dt $$

Now,

  $$ \int{\dfrac{dt}{{{t}^{2}}}} $$

 $$ =\int{{{t}^{-2}}dt} $$

 $$ =\dfrac{{{t}^{-2+1}}}{-2+1} $$

 $$ =-\dfrac{1}{t}=-\dfrac{1}{\cos x+\sin x} $$

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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