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Compute the indefinite integral $$ \int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx $$

My Attempt:

First, convert $$ \frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right) $$

This changes the integral to $$ \int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx $$

Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes

$$ \begin{align} \int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\ &= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\ &= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\ &= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right| \end{align} $$

where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.

Is this solution correct? Is there another method for finding the solution?

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  • $\begingroup$ You could also use $$\cos(2t)=\cos^2(t)-\sin^2(t)$$? $\endgroup$ – Chinny84 Mar 19 '14 at 19:28
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Let

\begin{equation*} I=\int \cos 2x\cdot\ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|\,dx. \end{equation*}

Using the following identity

\begin{equation*} \cos 2x=2\cos ^{2}x-1 \end{equation*}

and the substitution

\begin{equation*} u=\cos x, \end{equation*}

we get

\begin{equation*} I=\int \frac{1-2u^{2}}{\sqrt{1-u^{2}}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|\,du. \end{equation*}

$I$ is integrable by parts, differentiating the factor $\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|$ and integrating the factor $\frac{1-2u^{2}}{\sqrt{1-u^{2}}}$. After simplifying, we obtain

\begin{eqnarray*} I &=&u\sqrt{1-u^{2}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|+2\int \frac{u}{2u^{2}-1}du \\[2ex] &=&u\sqrt{1-u^{2}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|+\frac{1}{2} \ln \left| 2u^{2}-1\right| +C \\[2ex] &=&\left( \cos x\cdot\sin x\right)\cdot \ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|+\frac{1 }{2}\ln \left| 2\cos ^{2}x-1\right| +C\\[2ex] &=&\frac{\sin 2x }{2} \ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|+\frac{\ln \left| \cos 2x\right| }{2} +C. \end{eqnarray*}

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Integrate by parts: $\int udv=uv-\int v du$, where $$u=\ln\frac{\cos x+\sin x}{\cos x-\sin x}\Rightarrow du=\frac{\frac{(\cos x-\sin x)(-\sin x+\cos x)-(\cos x+\sin x)(-\sin x +\cos x) }{(\cos x-\sin x)^2}}{\frac{\cos x+\sin x}{\cos x-\sin x}}=...=\frac{2}{\cos 2x}dx$$ and $$ dv=\cos 2x dx \Rightarrow v=\frac{1}{2}\sin 2x.$$ Then, $$\int \cos 2x \ln(\frac{\cos x+\sin x}{\cos x-\sin x}) dx=\frac{1}{2}\sin 2x \ln\frac{\cos x+\sin x}{\cos x-\sin x}-\int \frac{1}{2}\sin 2x \frac{2}{\cos 2x} dx=$$ $$=\frac{1}{2}\sin 2x\ln \frac{\cos x+\sin x}{\cos x-\sin x}-\int \tan 2x dx= $$ $$=\frac{1}{2}\sin 2x \cdot\ln\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)-\frac{1}{2}\ln|\sec 2x|+c. $$

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