maths - integration

[denian]

integrate

\int\frac{\sin^2 x}{e^x}
= \int\frac{1}{2e^x} - \frac{\cos2x}{2e^x} dx
= - \frac{1}{2}e^{-x} - \frac{1}{2}\int\frac{\cos 2x}{e^x} dx

im trying to use the integration by parts method, but seems like im stuck there, doesnt reach the answer :-

-\frac{e^{-x}}{2} + \frac{e^{-x}}{10} ( cos 2x - 2 sin 2x ) + c

[Dr Transport]

write \cos(2x) = \frac{e^{2ix}+e^{-2ix}}{2} then integrate......is should work out.

[denian]

i never come across that formula. can u explain?

[denian]

ok..... i already know how to prove it. but may i know how u come across that formula?

[Dr Transport]

it is the expansion of the trigonometric functions in terms of complex exponentials......

[cookiemonster]

It's known as the Euler Formula.
http://mathworld.wolfram.com/EulerFormula.html

e^{i\theta} = \cos{\theta} + i\sin{\theta}

And, using the even/oddness of cos/sin,

e^{-i\theta} = \cos{\theta} - i\sin{\theta}

Which can then be combined to solve for either the sine or cosine.

cookiemonster

[DeadWolfe]

I wouldn't expect that he'd be required to know anything about complex numbers for this...

[denian]

yeah, it doesnt. anyways thanks for the suggestion.

[denian]

bump bmp bump

[Shahil]

write \cos(2x) = \frac{e^{2ix}+e^{-2ix}}{2} then integrate......is should work out.

:confused:

Just outta interest, how do u integrate that? do you just integrate as per real numbers or are their any special complex techniques?

[Dr Transport]

just integrate with respect to x, the complex number i is a constant.....

[TMcCloskey]

Repeated application of integration-by-parts is the suggested route. Denian set up the problem correctly, now just needs to proceed carefully:

Let the last integral represent I_1, ie,

I_1 = \frac{1}{2}\int\frac{\cos 2x}{e^x} dx

and select

u = \cos 2x

dv = e^{-x} dx

Then

2I_1 = \int u dv = uv - \int v du

2I_1 = -e^{-x} \cos 2x - 2 \int e^{-x} sin 2x dx

Now repeat this process with the last integral in expression for I1, ie, let

I_2 = \int e^{-x} sin 2x dx

Calculate via integration-by-parts and the resulting expression for I_2 should yield a trailing integral which you'll reconize as a multiple of I_1. Substitute "I_1" for this integral and collect terms to solve for "I_1". Substitute this into the original expression and your done.

[NateTG]

\int\frac{\sin(x)^2}{e^x} dx =
\int e^{-x} \sin(x)^2 dx
Now
u=-x
du=-dx
and
(\sin(x))^2=(-\sin(-x)^2=(\sin(-x))^2
so
-\int e^u \sin(u)^2 du =
-(e^u \sin(u)^2 - \int e^{u} 2 \sin(u) \cos(u) du)=
-(e^u \sin(u)^2 - \int e^{u} 2 \sin(2u) du)=
-(e^u \sin(u)^2 + e^{u} \cos(2u) + \int e^{u} \cos(2u))=
-(e^u \sin(u)^2 + e^{u}(\cos^2(u) - \sin^2(u)) + \int e^{u} \cos(2u) du)=
-e^u \cos^2(u) - \int e^{u} \cos^2(u)du + \int e^{u} \sin^2(u) du =
-e^u \cos^2(u) - \int e^{u} du + 2 \int e^{u} \sin^2(u) du=
so we have
\int e^{u} \sin^2{u} du = e^u \cos^2(u) + \int e^{u} du - 2 \int e^{u} \sin^2{u} du
so
\int e^{u} \sin^2{u} du = \frac{e^u \cos^2(u) + e^u)}{3}
so
\int\frac{\sin(x)^2}{e^x} dx = -\int e^u \sin(u)^2 du = -\frac{e^u \cos^2(u) + e^u}{3} + C
\int\frac{\sin(x)^2}{e^x} dx = - \frac{\cos^2(x)+1}{3e^x} + C

Of course, you should check my work etc.

[Shahil]

just integrate with respect to x, the complex number i is a constant.....

Why does my brain just decide to switch of sometimes...

Stupid, stupid, stupid! Of course i is a constant!