Hello, I am seeking for help in solving this defined integral. I tried to solve it but I am having problems with substitution.

I can think of several difficulties you might be having. Please show your work so we can see where the problem lies. Also, we'll want to see why you think you are wrong.

One possible issue is that you could remove the cosine from the radical, but it would be easy to make a mistake with signs.
 
To do this one needs to write the integral over two intervals: \(\displaystyle \left[0,\tfrac{\pi}{2}\right]~\&~\left[\tfrac{\pi}{2},\pi\right]\)
\(\displaystyle \int_0^{\frac{\pi }{2}} {\cos (x)\sqrt {({{\sin }^2}(x) + 1)} dx} + \int_{\frac{\pi }{2}}^\pi { - \cos (x)\sqrt {({{\sin }^2}(x) + 1)} dx} \)
 
Dr.Peterson said:
I can think of several difficulties you might be having. Please show your work so we can see where the problem lies. Also, we'll want to see why you think you are wrong.

One possible issue is that you could remove the cosine from the radical, but it would be easy to make a mistake with signs.
Click to expand...
So if make a substitution sinx=t, then i could "remove" the cosx, since cosxdx=dt, and dx=dt/cosx. However, that is if we pretend sqrt(cosx)^2 is cosx, and not apsolute cosx. But there is a problem with substitution with a defined integral, so with sinx=t, pi turns into 0, and 0 turns into 0. So this doesn't seem right to me. I don't know what i should try. The thing is, i would solve this one if it was a indefinite integral.
 
pka said:
To do this one needs to write the integral over two intervals: \(\displaystyle \left[0,\tfrac{\pi}{2}\right]~\&~\left[\tfrac{\pi}{2},\pi\right]\)
\(\displaystyle \int_0^{\frac{\pi }{2}} {\cos (x)\sqrt {({{\sin }^2}(x) + 1)} dx} + \int_{\frac{\pi }{2}}^\pi { - \cos (x)\sqrt {({{\sin }^2}(x) + 1)} dx} \)
Click to expand...
Thank you, but how did you get the minus on the second integral? Shouldn't it be just plus?
 
*For my previous message*
I just realised it's because of the absolute value. I understand now, thank you a lot @pka
 
stefan3070 said:
So if make a substitution sinx=t, then i could "remove" the cosx, since cosxdx=dt, and dx=dt/cosx. However, that is if we pretend sqrt(cosx)^2 is cosx, and not apsolute cosx. But there is a problem with substitution with a defined integral, so with sinx=t, pi turns into 0, and 0 turns into 0. So this doesn't seem right to me. I don't know what i should try. The thing is, i would solve this one if it was a indefinite integral.
Click to expand...
As I think you now realize, pka's hint showed how to prevent what I called a likely "mistake with signs" (namely, forgetting about the absolute value), and how to deal with that absolute value itself (by splitting into two parts). This also turns out to deal with the problem you asked about here, namely what to do with the limits of integration.

There's still more work to do, of course. Let us know if you get stuck on any subsequent step.
 
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