Can some one provide me hint to evaluate the following integral.
$$ \int\csc^{2}\left(x\right) \ln\left(\cos\left(x\right) - \sqrt{\vphantom{\largeA}\,\cos\left(2x\right)\,}\,\right) \,{\rm d}x $$.
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what is $cosec$? – Rustyn Apr 06 '14 at 07:59
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@Rustyn: I guess $\text{cosec}:x = \frac{1}{\sin x}$ (Cosecant) – user35603 Apr 06 '14 at 08:01
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Common notation is $\csc$, I guess I could be wrong though. My bad. – Rustyn Apr 06 '14 at 08:02
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@Rustyn: In some old books $\text{cosec}$ is used too. – user35603 Apr 06 '14 at 08:03
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@user35603 Ah, I see. Thanks. – Rustyn Apr 06 '14 at 08:03
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In fact with some trigonometry, we can simplify the integrand, in the formula of David H. (after the integration by parts step): $$ \cot x \cdot\frac{-\sin x+\frac{\sin 2x }{\sqrt{\cos 2x }}}{\cos{ x}-\sqrt{\cos 2x }}= \frac{\cos x}{\sin^2x\cdot\sqrt{1-2\sin^2 x}}+\cot^2x $$ This yields the following $$ \int \csc^2(x)\log(\cos{ x}-\sqrt{\cos 2x }) dx=-\cot x\cdot\log(\cos{x}-\sqrt{\cos 2x })-\cot x-x-\frac{\sqrt{\cos 2x}}{\sin x} $$
Omran Kouba
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Hint: integration by parts is a great candidate for a first step here. Using $\int\csc^2{x}\,dx=-\cot{x}$ and $$\frac{d}{dx}\log{\left(\cos{x}-\sqrt{\cos{2x}}\right)}=\frac{-\sin{x}+\frac{\sin{2x}}{\sqrt{\cos{2x}}}}{\cos{x}-\sqrt{\cos{2x}}},$$
we arrive at:
$$\int\csc^2{x}\log{\left(\cos{x}-\sqrt{\cos{2x}}\right)}dx\\ =-\cot{x}\log{\left(\cos{x}-\sqrt{\cos{2x}}\right)}-\int\cot{x}\cdot\frac{\sin{x}-\frac{\sin{2x}}{\sqrt{\cos{2x}}}}{\cos{x}-\sqrt{\cos{2x}}}dx.$$
David H
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I corrected a typo in the formula for the derivative of the logarithmic term. Presumably that was the reason for downvote? – David H Apr 06 '14 at 08:27
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MAPLE returns $$-\cot(x)\ln(\cos(x)-\sqrt{\cos(2x)})-\cot \left( x \right) +x-\sqrt { \left( -2\, \left( \cos \left( x \right) \right) ^{2}+1 \right) \left( \left( \cos \left( x \right) \right) ^{2}-1 \right) } \left( -1/2\,{\frac {- 2\, \left( \cos \left( x \right) \right) ^{3}+2\, \left( \cos \left( x \right) \right) ^{2}+\cos \left( x \right) -1}{\sqrt { \left( -\cos \left( x \right) -1 \right) \left( 2\, \left( \cos \left( x \right) \right) ^{3}-2\, \left( \cos \left( x \right) \right) ^{2}-\cos \left( x \right) +1 \right) }}}+1/2\,{\frac {-2\, \left( \cos \left( x \right) \right) ^{3}-2\, \left( \cos \left( x \right) \right) ^{2} +\cos \left( x \right) +1}{\sqrt { \left( 1-\cos \left( x \right) \right) \left( 2\, \left( \cos \left( x \right) \right) ^{3}+2\, \left( \cos \left( x \right) \right) ^{2}-\cos \left( x \right) -1 \right) }}} \right) \left( \sin \left( x \right) \right) ^{-1}{ \frac {1}{\sqrt {2\, \left( \cos \left( x \right) \right) ^{2}-1}}}$$
It seems you have a tricky integral on your hands. If I find a substitution that returns this I'll post
Millardo Peacecraft
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1Mathematica returns the far simpler result $$-\csc x\bigg{\sqrt{\cos2x}+x\sin x+\cos x\Big[1+\ln!\Big(!!\cos x-\sqrt{\cos2x}\Big)\Big]\bigg}.$$ – Lucian Apr 06 '14 at 09:01