Evaluating Trigonometric Integral

[jdawg]

Homework Statement

[itex]\int[/itex]^{[itex]\frac{\pi}{2}[/itex]}₀ sin⁷y dy
The bounds are from [itex]\frac{\pi}{2}[/itex] to 0.

Homework Equations

The Attempt at a Solution

I think I did the integration correctly, but I don't really know how to evaluate this.

[itex]\int[/itex]siny(sin²y)³ dy
[itex]\int[/itex]siny(1-cos²y)³ dy
u=cosy
du=-siny dy

-[itex]\int[/itex](1-u²)³ du
-[itex]\int[/itex](-u⁶+3u⁴-3u³+1) du

[itex]\frac{u⁷}{7}[/itex]-[itex]\frac{3u⁵}{5}[/itex]+[itex]\frac{3u⁴}{4}[/itex]-u

[itex]\frac{cos⁷y}{7}[/itex]-[itex]\frac{3cos⁵y}{5}[/itex]+[itex]\frac{3cos⁴y}{4}[/itex]-cosy

I'm a little rusty on my trig! Please help :)

 

[Ray Vickson]

Homework Statement

[itex]\int[/itex]^{[itex]\frac{\pi}{2}[/itex]}₀ sin⁷y dy
The bounds are from [itex]\frac{\pi}{2}[/itex] to 0.

Homework Equations

The Attempt at a Solution

I think I did the integration correctly, but I don't really know how to evaluate this.

[itex]\int[/itex]siny(sin²y)³ dy
[itex]\int[/itex]siny(1-cos²y)³ dy
u=cosy
du=-siny dy

-[itex]\int[/itex](1-u²)³ du
-[itex]\int[/itex](-u⁶+3u⁴-3u³+1) du

[itex]\frac{u⁷}{7}[/itex]-[itex]\frac{3u⁵}{5}[/itex]+[itex]\frac{3u⁴}{4}[/itex]-u

[itex]\frac{cos⁷y}{7}[/itex]-[itex]\frac{3cos⁵y}{5}[/itex]+[itex]\frac{3cos⁴y}{4}[/itex]-cosy

I'm a little rusty on my trig! Please help :)

No the bounds are not from ##\pi/2## to 0, they are from 0 to ##\pi/2##. Anyway, you have the right idea; you just need to figure out the bounds after the u-substitution; that is, what are ##a## and ##b## in
[tex] \int_0^{\pi/2} \sin^7 y \: dy = -\int_a^b (1-u^2)^3 \, du?[/tex]

 

[jdawg]

Oops! That's what I meant. Haha thanks, I guess it would be a lot easier to just change the bounds.

 

[jdawg]

Hey! I ran into another problem, I changed the bounds to 1 to 0:

-∫ (-u⁶+3u⁴-3u³+1) du

Then I flipped the bounds to 0 to 1 and changed the sign on the integral:

∫(-u⁶+3u⁴-3u³+1) du

Then I plugged in my bounds and got 99/140, but the answer in the back of the book is 16/35. I'm having trouble figuring out what I did wrong.

 

[Saitama]

Hey! I ran into another problem, I changed the bounds to 1 to 0:

-∫ (-u⁶+3u⁴-3u³+1) du

Then I flipped the bounds to 0 to 1 and changed the sign on the integral:

∫(-u⁶+3u⁴-3u³+1) du

Then I plugged in my bounds and got 99/140, but the answer in the back of the book is 16/35. I'm having trouble figuring out what I did wrong.

How do you get ##u^3##? It should be ##u^2## instead.

 

[jdawg]

How do you get ##u^3##? It should be ##u^2## instead.

Ooops! It was just a silly mistake. Thanks so much!