A reduction formula/integration by parts problem?

Question

If I (n) = Integral (limits pi/2 and 0) e^2x * Sin^n (x) dx

how do i show 2*I (n) = e^pi - n* Intergal (same limits) e^2x * Sin^(n-1) (x) cos (x) dx

Hence, prove

(4+n^2) I (n) = 2e^pi + n(n-1) * Intergrel(version n-2)

I can't quite get there although I have come close. Should I use reduction forumla method first or should i follow the intergration by parts first?

Any help will be appreciated!

Answer 1

Ouch

For the first part,
I(n) = INT(e^2x.(sin x)^n dx){pi/2,0}
u = (sin x)^n
du/dx = n.(sin x)^n-1.cos x
dv/dx = e^2x
v = 1/2e^2x

So with "uv - INT(v.du/dx dx){...}" (parts)

I(n) = [1/2e^2x.(sin x)^n]{pi/2,0} - INT(1/2e^2x.n.(sin x)^n-1.cos x dx){pi/2,0}

I(n) = 1/2e^pi - 1/2n.INT(e^2x.(sin x)^n-1.cos x dx){pi/2,0}
2I(n) = ...

Tada!
Can't do the other bit though, nor can I see how it could be done.

Answer 2

The two previous posters did the first part. For the second part, do *another* integration by parts with dv=e^2x and u=sin^(n-1)x cos x.
The derivative of u is then:
(n-1) sin^(n-2) x cos^2 x -sin^n x.

Now write cos^2 x=1- sin^2 x to see that du is
(n-1)sin^(n-2) x - n sin^n x.

Now, v=(1/2)e^2x ,so the term from uv in the integration by parts goes to 0 between the two limits since u is 0 at both limits. Plugging in, we now get
2I(n)=e^pi -n[ -(1/2) (n-1) I(n-2) + (1/2) n I(n)]
If you multiply by 2 and take the terms with I(n) to the other side, you get your result.

Answer 3

A point worth noting here is that whenever integrating functions of the form e^kx sin(x), after you have integrated by parts twice you end up with the integral you started with. So instead of integrating by parts again you replace this integral with I and rearrange for I.

I think this also happens with e^kx cos(x) but I'm not 100% sure.

Answer 4

Reduction Formula Integration By Parts

Answer 5

(| is integral sign )

I(n) = | e^2x * sin^n(x) dx

I(n) = 1/2 | sin^n(x) d e^2x ; ( de^2x = 1/2*e^2x dx )

I(n) = 1/2*sin^n(x)*e^2x /0,pi/2 - 1/2 | e^2x d sin^n(x)

now : AAA : | e^2x dsin^n(x) = | nsin^n-1()xcos(x) dx

thus
I(n) = 1/2* ( sin^n(pi/2)*e^pi - sin^n(0)*e^0 ) +AAA
I(n) = 1/2*e^pi + AAA. finished.