# Find `int_0^4 <sqrt(2t+1), sin(pi*t)> dt`

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To solve this integral, we simply treat it as two separate integrals:

`int_0^4 sqrt(2t+1) dt` and `int_0^4sin(pit) dt`

Let's solve the first integral first. To do this, let `u = 2t+1` giving us `(du)/2 = dt`. Let's make the substitution, rearranging the bounds based on u(t):

`1/2 int_1^9 sqrtu du`

Let's evaluate this integral by recognizing that `sqrtu = u^(1/2)` and that the integral can be evaluated pretty much like any polynomal term.

`1/2(2/3u^(3/2))` evaluated from u = 1 to u = 9:

`1/2*2/3 (9^(3/2) - 1^(3/2))`

Let's simplify:

`1/3(27 - 1) = 26/3`

We have solved the first integral! Now, let's move on to the second.

The easiest way to do this is to recognize that it is a periodic function with an average value of zero. Because we're moving over 2 complete periods of the function, we don't even have to do much math. The integral must be zero!

However, let's act like the problem was not so easy, so we'll evaluate the integral.

`int_0^4sin(pit) dt`

You can follow the link below to find that the integral of this function is:

`-1/pi cos(pit)` evaluated from t = 0 to t = 4:

`-1/pi(cos(4pi) - cos(0)) = -1/pi(1 - 1) = 0`

So, either way we choose to evaluate it the second integral is zero.

This gives our fully-evaluated vector integral:

`<26/3, 0>`

I hope this helps!

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