Mathematics

# Evaluate$\int \dfrac{dx}{\cos x \cos 2x}$

##### SOLUTION

We have,

$\int{\dfrac{dx}{\cos x\cos 2x}}$

On multiply and divide by $\cos x$

Now,

$\int{\dfrac{\cos xdx}{{{\cos }^{2}}x\cos 2x}}$

$\Rightarrow \int{\dfrac{\cos xdx}{\left( 1-{{\sin }^{2}}x \right)\left( 1-2{{\sin }^{2}}x \right)}}$

Let

$\sin x=t$

$\Rightarrow \cos xdx=dt$

$\Rightarrow \int{\dfrac{dt}{\left( 1-{{t}^{2}} \right)\left( 1-2{{t}^{2}} \right)}}$

Using partial dfraction and we get,

$\Rightarrow \int{\dfrac{dt}{\left( 1-{{t}^{2}} \right)\left( 1-2{{t}^{2}} \right)}}=\int{\left[ \dfrac{-1}{1-{{t}^{2}}}+\dfrac{2}{1-2{{t}^{2}}} \right]dt}$

On integrating and we get,

$\dfrac{1}{2}\ln \left( t-1 \right)\left( t+1 \right)+\dfrac{1}{\sqrt{2}}\ln \left( \dfrac{1+\sqrt{2}t}{1-\sqrt{2}t} \right)+C$

Put $t=\sin x$

So,

$=\dfrac{1}{2}\ln \left( \sin x-1 \right)\left( \sin x+1 \right)+\dfrac{1}{\sqrt{2}}\ln \left( \dfrac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right)$

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
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Chapters 126
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