Question

integral from 0 to pi/2 of sin^5 x cos^12 x dx without using the reduction formula

Answer 1

@SithsAndGiggles you did good in the last one, i did get it solved, not by the way you did it with half-angle formula but i used your input to check my answer. thank you.

Answer 2

\[\int_0^{\pi/2}\sin^5x\cos^{12}x\;dx\]
I would start by
\[\begin{align*}\sin^5x&= \sin x \sin^4x\\
&=\sin x (\sin^2x)^2\\
&=\sin x (1-\cos^2x)^2\end{align*}\]
Expand that, and you get something like
\[\int_0^{\pi/2}\left(\sin x \cos^{12}x - 2\sin x \cos^{14}x + \sin x\cos^{16}x\right)\;dx\]
(I didn't double-check that, so it may be wrong)
Then a simple u-sub shouldn't be too difficult.

Answer 3

Technically you can do this in the same way that the reduction formula is derived by going through a very long and tedious integration by parts. I'm trying to figure out if there's some clever way to cancel it out with one sweeping u-substitution or something since the derivative of sine is just cosine.

Yeah that looks pretty alright to me.

Answer 4

yeah, @Kainui, that is what i would like to use, but my prof. wants us to use the very long and tedious integration by parts... that is how it will be on the exam in a week or so

Answer 5

@SithsAndGiggles, good that is were I started. I'm going to keep working and see what happens

Answer 6

Did you say you wanted to do this by parts? I don't think there's any good way to do that.

Answer 7

Actually, what I meant to say was U-sub.

Answer 8

Just to make sure does \[(1-\cos^2 x)^2 = (1-2\cos^4 x+\cos^4 x) \]?

Answer 9

Close, the power of the second cosine term should be 2.

Answer 10

\[(1−2\cos^2x+\cos^4x)\] ?
Now also how did you expand from here? I know you plug in the \[sinx(1−\cos^2x)^2 \] for \[\sin^5x\]
but how to expand from there?

Answer 11

oh
haha nevermind. sorry.

Answer 12

well, @SithsAndGiggles what do you suggest i use for u-sub, just using sinx and break it down multiple times?

Answer 13

Let u = cosx, -du = sinx dx.
Then you have
\[\left(-\int_1^0u^{12}du\right) + \left(-\int_1^0u^{14}du\right) + \left(-\int_1^0u^{16}du\right)\\
\int_0^1u^{12}du+\int_0^1u^{14}du+\int_0^1u^{16}du\]

Answer 14

wouldn't the \[\int\limits u^{12}\] be negative?

Answer 15

Well here is the other thing too... This is a definite integral too... From 0 to pi/2 but I'm trying to solve the indefinite one first then deal with plugging in the numbers after

Answer 16

Sorry, I'm juggling a few things at a time here. Are you confused about the limits on the integrals? By changing integration over [1,0] to [0,1], the sign changes.

Answer 17

\[\int\limits -u^{12}+2u^{14}-u^{16}du\]

Answer 18

indeed, but if i dont change the limits then i can set it like this?

Answer 19

\[[-\frac{ u^{13} }{ 13 }+\frac{ 2u^{15} }{ 15 }-\frac{ u^{17} }{ 17 }]\]

Answer 20

Uh, let me rewrite what I've done. I feel like I'm tripping myself up a bit here...
\[\int_0^{\pi/2}\left(\sin x\cos^{12}x - \sin x\cos^{14}x + \sin x\cos^{16}x\right)dx\\
\int_0^{\pi/2}\sin x\cos^{12}x\;dx - \int_0^{\pi/2}\sin x\cos^{14}x\;dx + \int_0^{\pi/2}\sin x\cos^{16}x\;dx\]
After the sub, we get
\[\left(-\int_1^0u^{12}\;du\right) - \left(-\int_1^0u^{14}\;du\right) + \left(-\int_1^0u^{16}\;du\right)\\
\int_0^1u^{12}\;du-\int_0^1u^{14}\;du + \int_0^1u^{16}\;du\]
So it looks like I had a single sign mistake. Do you follow this?

Answer 21

Yeah, that is where I'm at right now. I am subbing in cosx for u right now and then solving from 0 to pi/2.

Answer 22

oh my - i came out to 0... that doesnt seem right

Answer 23

Well, before you do that (over 0 to pi/2, I mean), make sure you sub back u = cosx. Otherwise you'll get a wrong answer. I also left out a constant 2 on the second integral... Sorry about that.

Answer 24

Sooo this is what i just had written down...
\[[-\frac{ \cos^{13}x }{ 13 }+\frac{ 2\cos^{15}x }{ 15 }-\frac{ \cos^{17}x }{ 17 }]\int\limits_{0}^{\pi/2}\]
now that is not integral that is just meaning to plug in the endpoints into the equation.

Answer 25

When I just plugged this into my calculator, I came out with 0 though... I also tried plugging it into my homework and it did not accept it.

Answer 26

I guess the answer should be 0.002413273 when using the solver on my calculator.

Answer 27

Via the FTC, that last expression you have becomes
\[-\frac{1}{13}\left(\cos^{13}\frac{\pi}{2} - \cos^{13}0\right)+\frac{2}{15}\left(\cos^{15}\frac{\pi}{2} - \cos^{15}0\right)-\frac{1}{17}\left(\cos^{17}\frac{\pi}{2} - \cos^{17}0\right)\]
Which is equal to
\[-\frac{1}{13}(0-1)+\frac{2}{15}(0-1)-\frac{1}{17}(0-1)\\
=\frac{1}{13}-\frac{2}{15}+\frac{1}{17}\\
=\frac{8}{3315}\\
\approx0.00241\]

Answer 28

I feel so silly, when I plugged it into the calc I only did pi/2 i forgot that cos0 =1 so it wouldn't end up being all zero'd out for the second part... Thank you for walking through this with me.

Answer 29

I don't know if I was thinking sinx or what. either way - very good job @SithsAndGiggles !!!

Answer 30

Thanks :3