Evaluating Trigonometric Integral

[jdawg]

Homework Statement

$\int$^{$\frac{\pi}{2}$}₀ sin⁷y dy
The bounds are from $\frac{\pi}{2}$ 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.

$\int$siny(sin²y)³ dy
$\int$siny(1-cos²y)³ dy
u=cosy
du=-siny dy

-$\int$(1-u²)³ du
-$\int$(-u⁶+3u⁴-3u³+1) du

$\frac{u⁷}{7}$-$\frac{3u⁵}{5}$+$\frac{3u⁴}{4}$-u

$\frac{cos⁷y}{7}$-$\frac{3cos⁵y}{5}$+$\frac{3cos⁴y}{4}$-cosy

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

 

[Ray Vickson]

Homework Statement

$\int$^{$\frac{\pi}{2}$}₀ sin⁷y dy
The bounds are from $\frac{\pi}{2}$ 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.

$\int$siny(sin²y)³ dy
$\int$siny(1-cos²y)³ dy
u=cosy
du=-siny dy

-$\int$(1-u²)³ du
-$\int$(-u⁶+3u⁴-3u³+1) du

$\frac{u⁷}{7}$-$\frac{3u⁵}{5}$+$\frac{3u⁴}{4}$-u

$\frac{cos⁷y}{7}$-$\frac{3cos⁵y}{5}$+$\frac{3cos⁴y}{4}$-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
$$\int_0^{\pi/2} \sin^7 y \: dy = -\int_a^b (1-u^2)^3 \, du?$$

 

[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!