Question

Someone help me solve this integral?

Answer 1

\[\int\limits \sin^5\cos^2dx\]

Answer 2

As the saying goes, take the odd man out!

Answer 3

right, so it becomes \[\int\limits \sin^4(x) * \sin(x) * \cos^2(x) dx\]

Answer 4

Yup, keep going

Answer 5

Hrmmmmmmm. I guess we could say sin^2(x)^2?

Answer 6

You pro!

Answer 7

hahaha. i see it now... we can say by sin^2x + cos^2x = 1 that sin^2x = 1+cos^2x right?

Answer 8

and sub that for sin^2(x)?

Answer 9

Careful, sin^2x = 1-cos^2x

Answer 10

Oops yeah minus sorry

Answer 11

You got this!

Answer 12

so it becomes \[\int\limits [(1-\cos^2x)^2 * \sin(x) *\cos^2(x)dx\]

Answer 13

Right

Answer 14

now im lost cause I'm not sure of what next step to take... are we going to use a trig identity here or?

Answer 15

Nope, a u substitution :)

u = cosx

Answer 16

OH

Answer 17

hehe

Answer 18

i see. u = cos x, du = -sinx so:

\[\int\limits (1-u^2)^2 * u^2*-du \]

Answer 19

so we just pull the negative outta du

Answer 20

Yes if you like :)

Answer 21

Then we foil and take the antiderivative?

Answer 22

Yup, simple algebra now

Answer 23

niceeeeee. it was just the u-sub. how do you really know when to use it?

Answer 24

The thing is, with integrals, the first line of attack is usually a u substitution. But we notice that there is only 1 sinx, and you have cosx's flying around and we know of course the derivative of cosx is -sinx. But usually with these trig questions it requires a u sub

Answer 25

Let me give you some rules

Answer 26

alright

Answer 27

oh my, thats beautiful

Answer 28

And a link http://puu.sh/mXqtr/d091ec0f05.png

Answer 29

So you can save it haha

Answer 30

thanks! physics/calc god hahaha

Answer 31

Yw xD