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.

  • $\begingroup$ Did you try taking the derivative of the function on the right? $\endgroup$
    – LL 3.14
    May 16, 2020 at 4:44

2 Answers 2


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}



$$\begin{align} \left(\arcsin\frac{y+\sin y}{1+y\sin y}\right)'=\frac{1-y^2+\cos y}{(1+y\sin y)\sqrt{1-y^2}} \end{align}$$ Then let y=sinx, Problem solved.


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