How do you evaluate the definite integral #int 1+sinx# from #[pi/4, pi/2]#?

Question

Christopher Agresta · Accepted Answer

#pi/4+sqrt2/2# #int_(pi/4)^(pi/2)(1+sinx)dx# #=[x-cosx]_(pi/4)^(pi/2)# #=[x-cosx]^(pi/2)-[x-cosx]_(pi/4)# #(pi/2-cos(pi/2))-(pi/4-cos(pi/4))# #=pi/2-0-pi/4+sqrt2/2# #pi/4+sqrt2/2#

Adam Adkins · Answer

undefined To evaluate the definite integral $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (1 + \sin(x)) \, dx$, you can proceed as follows:

1. Integrate $1$ with respect to $x$ within the given limits:
   $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \, dx = x \Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$

2. Integrate $\sin(x)$ with respect to $x$ within the given limits:
   $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x) \, dx = -\cos(x) \Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{4}\right) = -0 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$$

3. Add the results from step 1 and step 2:
   $$\frac{\pi}{4} + \frac{\sqrt{2}}{2}$$

So, $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (1 + \sin(x)) \, dx = \frac{\pi}{4} + \frac{\sqrt{2}}{2}$$