indefinite-integralsintegrationtrigonometric-integrals
From software, I get
$$\int \frac{\cos ^2 x+\cos (\sin x)}{\sin x \sin (\sin x)+1} \, dx=-2 \tan ^{-1}\left(\cos \left(\frac{x}{2}-\frac{\sin x}{2}\right) \csc \left(\frac{x}{2}+\frac{\sin x}{2}\right)\right)$$
I tried $t=\sin x,x=\arcsin t$ and get
$$\int \frac{1-t^2+\cos t}{\sqrt{1-t^2} (1+t \sin t)} \, dt$$
Then, I can't figure out how to do it.
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*}
$\alpha = \cot^{-1} x\\ \beta = \tan^{-1} (\sin \alpha)$
Use trig indentities to find $\csc \alpha$ and $\sin \alpha$
$\sin\alpha = \tan \beta$
Use similar identities to find $\sec \beta$ and $\cos\beta$
Best Answer
Rewrite the denominator
\begin{align} \sin x \sin (\sin x)+1 = &\frac12[\cos(x-\sin x) - \cos(x+\sin x)]+1 \\ = &\frac12[1-\cos(x+\sin x)] + \frac12[1+\cos(x-\sin x)] \\ = & \sin^2t_+ + \cos^2t_- \end{align} where $t_\pm=\frac x2 \pm \frac{\sin x}2$. Also rewrite the numeritor
\begin{align} \cos ^2x+\cos (\sin x)= &\cos x\cos(t_+ + t_-) + \cos(t_+ - t_-) \\ = &(1+\cos x)\cos t_+ \cos t_- + (1-\cos x)\sin t_+ \sin t_- \\ = & 2 \cos t_- \frac{d\sin t_+}{dx} - 2\sin t_+ \frac{d\cos t_-}{dx} \end{align}
Substitute above into the integrand to obtain
\begin{align} \int \frac{\cos ^2 x+\cos (\sin x)}{\sin x \sin (\sin x)+1} \, dx & =\int \frac {2 \cos t_- \frac{d\sin t_+}{dx} - 2\sin t_+ \frac{d\cos t_-}{dx} }{ \sin^2t_+ + \cos^2t_-} \\ & =-2\int \frac {\sin^2 t_+ \frac{d}{dx}\left(\frac{\cos t_-}{\sin t_+}\right) }{ \sin^2t_+ + \cos^2t_-} \\ & =-2\int \frac {\frac{d}{dx}\left(\frac{\cos t_-}{\sin t_+}\right) }{ 1 + \left(\frac{\cos t_-}{\sin t_+}\right)^2} = -2\tan ^{-1}\left(\frac{\cos t_-}{\sin t_+}\right) +C \end{align}