Question

Integral from -pi/2 to pi/2 ( 16 + 24sinx + 9sin^2(x) ) ?
always MEDAL

Answer 1

In the solution manual I don't get why it cities about odd and even functions then magically, the middle term (24sinx) goes away with the problem exactly the same.

Answer 2

\[1/2\int\limits_{-\pi/2}^{\pi/2}16 +24sinx + 9\sin ^{2}x dx\]

Answer 3

I'm in the area of polar coordinates chapter.

Answer 4

Do not solve the integral! I'm just asking how the 24sinx went away.

Answer 5

@SolomonZelman

Answer 6

@wio

Answer 7

The only way I could think of it going away is integrating the portion \(\int_{\pi/2}^{\pi/2} 24\sin(x)\)

Answer 8

ya, actually I think they just didn't show that step

Answer 9

should've just continued and solved it

Answer 10

You'd be interested in this:
http://openstudy.com/study#/updates/548912eae4b0209d5c4e0cbc

Answer 11

16 + 24sinx + 9sin^2(x) = (4+3sin(x))(4+sin(x))

Answer 12

go ahead, lol

Answer 13

\[\large \int_{-\pi/2}^{\pi/2}16dx+\int_{-\pi/2}^{\pi/2} 24\sin(x)dx+\int_{-\pi/2}^{\pi/2}9\sin^2(x)\]

Answer 14

Jhanny, it is a perfect square trinomial.

Answer 15

Ooh it is!! I see that now.

Answer 16

\[\int\limits_{-\pi /2}^{\pi/2} ( 3 \sin x +4 ) ^2~dx\]

Answer 17

Actually, that doesn't necessarily simplify in this case.

Answer 18

Yeah it's not a simple enough function lol

Answer 19

You'd need a u sub and stuff :(

Answer 20

Maybe.

Answer 21

no, half angle identity

Answer 22

I don't know why the mayde it such a perfect square trinomial though. Im have not so much time right now, I wish i did....

Answer 23

cosine or sine, why does that make a difference, perhapsm doing it just term by term as you did is the best way.

Answer 24

I got to go, sorry guys.

Answer 25

\[\sin(2x) = \frac{1}{2} +\frac{1}{2}(1-\cos(2x))\]