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Evaluation of $\displaystyle \int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$

My Try:: Let $\displaystyle I = \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx = \int \ln(\cos x+\sqrt{\cos 2x})\cdot \csc^2 xdx$

So $\displaystyle I = -\ln\left(\cos x+\sqrt{\cos 2x}\right)\cdot \cot x+\int \frac{1}{\left(\cos x+\sqrt{\cos 2x}\right)}\cdot \left(-\sin x-\frac{1}{2\sqrt{\cos 2x}}\cdot 2\cdot \sin 2x\right)\cdot \cot xdx$

Now How Can I solve after that

Help me

Thanks

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As you have done, we first use integration by parts to get $$\begin{align} I &= \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx \\&= \int \ln(\cos x+\sqrt{\cos 2x})\csc^2 (x)dx \\&=-\cot (x)\ln(\cos x+\sqrt{\cos 2x})+\int \left(\frac{1}{\cos x+\sqrt{\cos 2x}}\right) \left(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}} \right)\cot (x)dx \end{align}$$

Now we just need to deal with the remaining integral which we will call $J$. First note that $$(\cos x+\sqrt{\cos 2x})(\cos x-\sqrt{\cos 2x})=\cos^2x-\cos2x=\cos^2x-(\cos^2x-\sin^2x)=\sin^2x$$

Hence $$\begin{align} J&=\int \left(\frac{1}{\cos x+\sqrt{\cos 2x}}\right) \left(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}} \right)\cot (x)dx \\&=\int\left(\frac{\cos x-\sqrt{\cos 2x}}{\cos x-\sqrt{\cos 2x}}\right)\left(\frac{1}{\cos x+\sqrt{\cos 2x}}\right) \left(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}} \right)\cot (x) dx \\&=\int\left(\frac{\cos x-\sqrt{\cos 2x}}{\sin^2x}\right) \left(-\sin x-\frac{2\sin x\cos x}{\sqrt{\cos 2x}} \right)\cot (x) dx \\&=\int\left(\frac{-\sin x\cos x+\sin x\sqrt{\cos 2x}-\frac{2\sin x\cos^2 x}{\sqrt{\cos2x}}+2\sin x\cos x}{\sin^2x}\right)\cot (x) dx \\&=\int\frac{-\cos^2 x+\cos x\sqrt{\cos 2x}-\frac{2\cos^3 x}{\sqrt{\cos2x}}+2\cos^2 x}{\sin^2x}dx \\&=\int\left(-\frac{\sin^2x-1}{\sin^2x}+\frac{\sin^2x-1}{\sin^2x}\right)+\frac{\cos^2x +\cos x\sqrt{\cos 2x}-\frac{2\cos^3 x}{\sqrt{\cos2x}}}{\sin^2x}dx \\&=-x-\cot x+\int\frac{\cos x\sqrt{\cos 2x}-\frac{2\cos^3 x}{\sqrt{\cos2x}}}{\sin^2x}dx \\&=-x-\cot x+\int\frac{\cos x\cos 2x-2\cos^3 x}{\sin^2x\sqrt{\cos2x}}dx \\&=-x-\cot x+\int\frac{\cos x(1-2\sin^2x)-2\cos^3 x}{\sin^2x\sqrt{\cos2x}}dx \\&=-x-\cot x+\int\frac{-2\cos x\sin^2x+\cos x(1-2\cos^2 x)}{\sin^2x\sqrt{\cos2x}}dx \\&=-x-\cot x+\int\frac{-\sin2x\sin x-\cos x\cos2x}{\sin^2x\sqrt{\cos2x}}dx \\&=-x-\cot x+\int\frac{\frac{-\sin2x}{\sqrt{\cos2x}}\sin x-\cos x\sqrt{\cos2x}}{\sin^2x}dx \\&=-x-\cot x+\frac{\sqrt{\cos 2x}}{\sin x}+C \end{align}$$

Hence

$$\begin{align}I&=-\cot (x)\ln(\cos x+\sqrt{\cos 2x})-x-\cot x+\frac{\sqrt{\cos 2x}}{\sin x}+C \\&=\frac{\sqrt{\cos 2x}}{\sin x}-x-\cot (x)(1+\ln(\cos x+\sqrt{\cos 2x})+C \end{align}$$

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  • $\begingroup$ I do just want to note that this solution was informed by the answer (I knew I wanted to manipulate the integral to get out an $x+\cot x$, and I knew I wanted to make the rest look like a quotient rule). $\endgroup$ – Peter Woolfitt Aug 12 '14 at 10:02
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Hint: After integrating by parts, try substituting $u=\sin{x}$. Also, it will be helpful to recall the trigonometric identity, $\cos{(2x)}=\cos^2{x}-\sin^2{x}=2\cos^2{x}-1=1-2\sin^2{x}$.

Let $I$ be your integral, then:

$$\begin{align} I &=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\cot{x}\left(\sin{x}+\frac{\sin{(2x)}}{\sqrt{\cos{(2x)}}}\right)}{\cos{x}+\sqrt{\cos{(2x)}}}\mathrm{d}x\\ &=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\cos{x}\left(1+\frac{2\cos{x}}{\sqrt{\cos{(2x)}}}\right)}{\cos{x}+\sqrt{\cos{(2x)}}}\mathrm{d}x\\ &=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\cos{x}\left(1+\frac{2\sqrt{1-\sin^2{x}}}{\sqrt{1-2\sin^2{x}}}\right)}{\sqrt{1-\sin^2{x}}+\sqrt{1-2\sin^2{x}}}\mathrm{d}x\\ &=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\left(1+\frac{2\sqrt{1-u^2}}{\sqrt{1-2u^2}}\right)}{\sqrt{1-u^2}+\sqrt{1-2u^2}}\mathrm{d}u\\ &=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\frac{1}{\sqrt{1-2u^2}}-\sqrt{1-u^2}}{u^2}\mathrm{d}u. \end{align}$$

This last integral may then be split up into two fairly easy integrals.

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  • $\begingroup$ Let me know if the edit is OK. I just edited to make your text fit in the page. $\endgroup$ – Dmoreno Aug 12 '14 at 10:13

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