Indefinite Integral Doubt!

Please help me solve the following Integral :

$\large I = \int\dfrac{\sqrt{cos(2x)}}{cosx} dx$

$Note$ : Anyone and everyone please post your solution , this question appeared in my class test.

#Calculus

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Comments

$\dfrac{\sqrt{\cos 2x}}{\cos x} = \dfrac{\cos 2x}{\cos x \sqrt{\cos 2x}}$

$\dfrac{2\cos^2 x -1}{\cos x \sqrt{2\cos^2 x -1}} = \dfrac{2\cos x }{\sqrt{2\cos^2 x -1}}-\dfrac{1}{\cos x \sqrt{2\cos^2 x -1}}$

For first term setting $\sin x = u$ will convert it into a standard integral. For the second term first set $\sin x = u$ then then set $\dfrac{1}{t}=u$ . This will convert it into a standard integral.

A Former Brilliant Member - 3 years, 7 months ago

Hey nice approach , thank you bro, i wonder if you are this good from self - study or do you go to some good coaching? $\ddot\smile$