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
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