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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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