I(n) = ∫sin^n (x) dx (integration by parts)

[vmr101]

Homework Statement

i)Use integration by parts to express:
I(n) = ∫ sin^n (x) dx
in terms of I(n-2).
ii) Hence show that ∫(π/2 for top, π/4 for bottom) 1/[sin^4 (x)] dx = 4/3

Homework Equations

Reduction Formula and Trig Identity [sin²(x) + cos²(x) = 1]
π = pi

The Attempt at a Solution

i) Integrate by parts with
u = sin^(n-1)(x) => du = (n-1)·cos(x)·sin^(n-2)(x)
dv = sin(x) => v = -cos(x)
I ended up with:
∫ sin^n(x) dx
= -(1/n)·cos(x)·sin^(n-1)(x) + [(n-1)/n]· ∫ sin^(n-2)(x) dx

ii) Would this be integration by substitution and parts?
= log{ -(1/4)·cos(x)·sin^(3)(x) + [3/4]· ∫ sin^(2)(x) dx }
=> log{ -(1/4)·cos(x)·sin^(3)(x) + [3/4]·(-(1/2)·cos(x)·sin(x) + [1/2]) }
I pushed the numbers through but got this wrong, where did i go wrong?
Thanks in advance.

 

[vela]

You correctly found

[tex]I(n) = -\frac{1}{n}\cos x\sin^{n-1} x + \frac{n-1}{n} I(n-2)[/tex]

Now you're asked to evaluate

[tex]\left \int_{\pi/4}^{\pi/2} \sin^{-4} x \, dx = I(-4)\right\rvert_{x=\pi/4}^{x=\pi/2}[/tex]

But it looks like you used n=4. Try a different value of n. Also, where did the log come from?

 

[vmr101]

Of course! Thanks.
I don't know why, but i used substitution to change everything under the 1 to u, then used the integral of 1/u is log, then replaced u back in, then used the 4.
I knew i was doing something wrong.
Thanks for the help! Its much appreciated.

 

[vmr101]

I put int 1/sin^4 into wolframa alpha website and got
-(1/3)*cot(x)(csc^2(x)+2)
I then put pi/2 and pi/4 and got the answer 4/3.
Is anyone able to explain the integral process for:
int 1/sin^4(x)
??
Thanks

 

[swerty]

You have found:

[tex]I(n) = -\frac{1}{n}\cos x\sin^{n-1} x + \frac{n-1}{n} I(n-2)[/tex]

And we know:

[tex]I(-2)=\int \frac{dx}{sin^2(x)}=-cot(x)+C[/tex]

Then for finding I(-4) you have to use the formula for n=-2.