Integrating Sin Squared: Solve \int\frac{\sin^2 x}{e^x} dx

[denian]

integrate
[tex]
\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
[/tex]
im trying to use the integration by parts method, but seems like I am stuck there, doesn't reach the answer :-

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

 

[Dr Transport]

write [tex] \cos(2x) = \frac{e^{2ix}+e^{-2ix}}{2} [/tex] 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

[tex]e^{i\theta} = \cos{\theta} + i\sin{\theta}[/tex]

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

[tex]e^{-i\theta} = \cos{\theta} - i\sin{\theta}[/tex]

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

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 [tex] x[/tex], the complex number [tex] i [/tex] 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 [itex]I_1[/itex], ie,

[tex]I_1 = \frac{1}{2}\int\frac{\cos 2x}{e^x} dx[/tex]

and select

[tex]u = \cos 2x [/tex]

[tex]dv = e^{-x} dx[/tex]

Then

[tex]2I_1 = \int u dv = uv - \int v du [/tex]

[tex]2I_1 = -e^{-x} \cos 2x - 2 \int e^{-x} sin 2x dx [/tex]

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

[tex]I_2 = \int e^{-x} sin 2x dx[/tex]

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

 

[NateTG]

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

Of course, you should check my work etc.

 

[Shahil]

just integrate with respect to [tex] x[/tex], the complex number [tex] i [/tex] is a constant...

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

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