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.

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## Comments

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TopNewest\(\dfrac{\sqrt{\cos 2x}}{\cos x} = \dfrac{\cos 2x}{\cos x \sqrt{\cos 2x}}\)

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

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.

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

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Self study. Although I have taken DLP from Resonance.

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@Md Zuhair , @Mayank Singhal , guys any help?

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Let me try ... :)

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