Question

I could really use a walk through on this: integrate cosxsin(sinx)dx from 0 to pi/2

Answer 1

I thought maybe a u substitution for sinx. mostly I have forgotten the trig and in the class we are not allowed the use of calculators.

Answer 2

Ooo interesting problem :O

Let's rearrange things a tiny bit.

int sin(sin x) * (cos x dx)

So you thought maybe u = sin x would be a good idea?
Yesssssssssss

Answer 3

If you take the derivative of that substitution, what do you get for du?

Answer 4

I get du = cos x dx

Answer 5

ok good good good.
So we get somethingggggg like this, yes?

int sin(sin x) * (cos x dx) ---> int sin(u) * (du)

Answer 6

yes that is exactly what I came up with!

Answer 7

k good.
Do you remember anti-derivative of sin? :U

Answer 8

-cos

Answer 9

cool.

= -cos(u)

We have bounds to evaluate this at?
Before we plug them in, we have to do something.

So we have a couple of options here.
We can either find `new boundaries` by plugging the old ones into our substitution.

Or we can `undo our subistution`.

Which method are you more comfortable with?

Answer 10

Our boundaries are written in terms of x=
No bueno.

We need boundaries for u.
Or undo.

Answer 11

This is where I am stuck...I do not remember this at all really

Answer 12

Personally I prefer undoing the substitution.
Let's go with that.

So our result was -cos(u).

Our substitution was u = sinx.

So let's plug that back in, giving us,

-cos(sin x)

Ok good we undid our sub, getting it back in terms of x.
Now we're allowed to use the boundaries of 0 and pi/2 ( since those were bounds on x).

Answer 13

Ok I had tried that but the problem I ran into was finding -cos(1) without a calculator.

Answer 14

Ya you would have to stop there. That would be part of your final answer.
You can't find that easily without a calculator.

Answer 15

oh good I took calc 1 years ago and was terrible with trig even then...so I thought there must be something I was missing.

Answer 16

Thanks so much for the help!

Answer 17

yah they're just trying to be sneaky :)
np!