# How to evaluate $\int_{0}^{\pi/2}\frac{\sin(x+\sin x)}{\sin x}\,dx$

How to evaluate $$I=\int_{0}^{\frac{\pi}{2}}\frac{\sin (x+\sin x)}{\sin x}\,dx$$ I tried to expand $$I=\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos (\sin x)+\cos x\sin (\sin x)}{\sin x}\,dx$$ $$=\int_{0}^{\frac{\pi}{2}}\left(\cos (\sin x)+\frac{\cos x\sin (\sin x)}{\sin x}\right)dx$$ $$=\int_{0}^{\frac{\pi}{2}}\cos (\sin x)\,dx+\int_{0}^{\frac{\pi}{2}}\frac{\sin (\sin x)}{\sin x}\,d(\sin x)$$ $$=\int_{0}^{\frac{\pi}{2}}\cos (\sin x)\,dx+\int_{0}^{1}\frac{\sin x}{x}\,dx$$ $$=\int_{0}^{\frac{\pi}{2}}\cos (\sin x)\,dx+\operatorname{Si}(1)$$ But I don't know what to do next, help me, thank you

• Check bessel's function. Jun 20, 2019 at 5:25
• Thank you very much!
– FofX
Jun 20, 2019 at 8:01

What you did is correct! Let me help you out a little bit. The bessel function's integral representation is given by $$J_n(z)=\frac1\pi \int\limits_{0}^{\pi}\cos(tn-z\sin t) dt$$ Thus, $$J_0(1)=\frac1\pi \int\limits_{0}^{\pi}\cos(-\sin t) dt$$ $$J_0(1)=\frac2\pi \int\limits_{0}^{\pi/2}\cos(\sin t) dt$$
which gives your result by replacing variables $$\int\limits_{0}^{\pi/2}\cos(\sin x) dx = \frac{\pi}{2}J_0(1)$$
So, the final asnwer of your integral becomes $$\boxed{\frac{\pi}{2}J_0(1)+ \text{Si}(1)}$$
• I was curious so I looked up the value: $2.148\,053$. The link also has some fun alternate representations. Jun 20, 2019 at 6:17