Question

Could use some help knowing where to get started on this one, it looks a bit daunting:
\[\int\limits_{0}^{\pi} (4x + 9 \sin x)^2 dx\]
(don't need to evaluate, that's the easy part; a best answer medal will be given)

Ty in advance for your time :-)



Previously unanswered questions today (now closed because I give up on waiting, medals still up for grabs):
http://openstudy.com/study#/updates/4ff1d3e7e4b03c0c488a8864
http://openstudy.com/study#/updates/4ff1c08fe4b03c0c488a70ac

Answer 1

use integration by parts

Answer 2

and then use u- substitution

Answer 3

\[\int (16 x^2+81 \sin^2(x)+72 x \sin(x)) dx\]

Answer 4

for the sin^2 bit use the trig identity \[\sin^2u=\frac12(1-\cos(2u))\]

Answer 5

u = x, dv = sin(x) dx,
du = dx, v = -cos(x)

Answer 6

Oh! The power reduction formula Turing!

Answer 7

yeah, gonna need at least the double angle formula, but you could also use power reduction (I hate it personally)

Answer 8

u = (4x+9sinX)^2
du = 2(4x+9sinx)(4+9cosx)dx
dv = 1
v = x

Answer 9

Turns into this, yes?
\[81 \int\limits \frac{1}{2} dx-\frac{81}{2} \int\limits \cos(2 x) dx\]

Answer 10

(I'm writing down on a scratchpad as I go)

Answer 11

yep

Answer 12

\[16 \int\limits x^2 dx-72 x \cos(x)+81 \int\limits 1/2 dx-81/2 \int\limits \cos(2 x) dx+72 \int\limits \cos(x) dx\]

Answer 13

Sry that was the whole thing :D

Answer 14

\[16 \int\limits x^2 dx-72 x \cos(x)+81 \int\limits 1/2 dx..\]\[..-81/2 \int\limits \cos(2 x) dx+72 \int\limits \cos(x) dx\]

Answer 15

Ah and then a u-sub...
u = 2x
du = 2 dx \(\rightarrow\) 1/2 du = dx

Answer 16

\[-81/4 \int\limits \cos(u) du+16 \int\limits x^2 dx-72 x \cos(x)+81 \int\limits 1/2 dx+72 \int\limits \cos(x) dx\]

Answer 17

Yeesh that's getting messy

Answer 18

take each part separately...

Answer 19

Still good?
\[-\frac{(81 \sin(u))}{4}+\frac{16 x^3}{3}+\frac{81 x}{2}+72 \sin(x)-72 x \cos(x)\]

Answer 20

Err +C :D

Answer 21

Forgot my C, sry gents

Answer 22

you still have some stuff with u in it, so I'm confused...

Answer 23

\[\int16x^2+72x\sin x+81\sin^2xdx\]\[\int16x^2+72x\sin x+\frac{81}2(1-\cos (2x))dx\]\[\int16x^2+\frac{81}2dx-\frac12\int\cos(2x)dx+72\int x\sin xdx\]yes I thin you answer is correct, but the u should be a 2x

Answer 24

\[\frac{16 x^3}{3}+\frac{81 x}{2}+72\sin(x)-\frac{81}{4} \sin(2 x)-72 x \cos(x) + C\]

Answer 25

Yeah that's what I had too, u = 2x

Answer 26

\[\int16x^2+72x\sin x+81\sin^2xdx\]\[\int16x^2+72x\sin x+\frac{81}2(1-\cos (2x))dx\]\[\int16x^2+\frac{81}2dx-\frac{81}2\int\cos(2x)dx+72\int x\sin xdx\]\[\frac{16}3x^3+\frac{81}2x-\frac{81}4\sin(2x)-72x\cos x+72\sin x+C\]looks like I get the same :D

Answer 27

sin(0) = 0
sin(\(\pi\)) = 0
cos(0) = 1
cos(\(\pi\)) = -1

Therefore...

Answer 28

\[\frac{675\pi}{6} + \frac{16 \pi^3}{3})\]

Answer 29

That makes sense, thank you both @Turing & @niravshah08 :-D

Answer 30

and wolfram agrees :)
http://www.wolframalpha.com/input/?i=integral+0+to+pi+of+%284x%2B9sinx%29%5E2dx
so we're good
onto the next if I have time, I have to go teach English soon lol

Answer 31

The answer should be around 520-ish. And it's ~518.8

Answer 32

Yay english!

Answer 33

(sorta)

Answer 34

Good luck with that :D