Integral calculation explanation please

[tapp]

Hello
I'm a bit rusty of integral techniques , can you tell how the integral in the following picture was calculated please?

You can ignore the first transition, only the second one is relevant.(from the sin +cos integral to the a dt part.

 

[pka]

You can ignore the first transition, only the second one is relevant.(from the sin +cos integral to the a dt part.

Just note that \(\displaystyle (-a\sin(t))^2+(a\cos(t))^2=(-a)^2\sin^2(t)+(a)^2\cos^2(t)\)

 

[tapp]

Just note that \(\displaystyle (-a\sin(t))^2+(a\cos(t))^2=(-a)^2\sin^2(t)+(a)^2\cos^2(t)\)

And then?
I guess we use the identity sin^2+cos^2=1 somehow?
But what happens to the (-a^2) and a^2?
Thanks

 

[pka]

But what happens to the (-a^2) and a^2?

First of all, it is \(\displaystyle (-a)^2\) NOT \(\displaystyle (-a^2)\). You should know that \(\displaystyle -a^2\ne(-a)^2~!!\)

 

[tapp]

Yes my bad spelling

 

[Ishuda]

Just to be pedantic
\(\displaystyle \sqrt{(-a\space sin(t))^2 + (a\space cos(t))^2} = \sqrt{(a\space sin(t))^2 + (a\space cos(t))^2}\)
\(\displaystyle = \sqrt{a^2 sin^2(t) + a^2 cos^2(t)}\)
\(\displaystyle = \sqrt{a^2 (sin^2(t) + cos^2(t))}\)
\(\displaystyle = \sqrt{a^2}\)
\(\displaystyle =\lvert{a}\rvert\)

Assuming a is non-negative \(\displaystyle \lvert{a}\rvert\) = a

 

[HallsofIvy]

Just to be pedantic
\(\displaystyle \sqrt{(-a\space sin(t))^2 + (a\space cos(t))^2} = \sqrt{(a\space sin(t))^2 + (a\space cos(t))^2}\)
\(\displaystyle = \sqrt{a^2 sin^2(t) + a^2 cos^2(t)}\)
\(\displaystyle = \sqrt{a^2 (sin^2(t) + cos^2(t))}\)
\(\displaystyle = \sqrt{a^2}\)
\(\displaystyle =\lvert{a}\rvert\)

Assuming a is non-negative \(\displaystyle \lvert{a}\rvert\) = a

And, being even more pedantic but this appears to be what the OP was asking, for any constant, a, \(\displaystyle \int_0^{2\pi} a dx= \left[ax\right]_0^{2\pi}= a(2\pi- 0)= 2\pi a\)