Mathematics

Solve $$\displaystyle\int {\dfrac{{\cos 2x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx$$


SOLUTION

Consider the given integral.

$$I=\int{\dfrac{\cos 2x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx$$

 

We know that

$$\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$$

 

Therefore,

$$I=\int{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx$$             ……… (1)

 

Let

$$t=\sin x\cos x$$

$$ \dfrac{dt}{dx}=\cos x\left( \cos x \right)+\sin x\left( -\sin x \right) $$

$$ dt=\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)dx $$

 

Therefore,

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

$$ I=-\dfrac{1}{t}+C $$

 

On putting the value of $$t$$, we get

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


Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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