calculus trigonometric, exponential integration by parts?

Question

Okay this is the probem:
int[(sin(x))*(e)^(2 x) dx
I understood it until the part i circled in green on this link:

http://us.a2.yahoofs.com/users/43db6743zfb206226/5a27re2/__sr_/ee3dre2.jpg?phICsTFBAnSDctOF


then I couldn't understand where the 5/4 came from(the part I circled in red. Please explain this to me. I missed the class where we were taught to do this so i probably missed something

Answer 1

Distribute the -1/2 through the parentheses and you get a
-1/4 integral e^2x sin x

Add this to both sides and you will get the 5/4 integral of e^2x sinx

Answer 2

For expressions involving products of e^x, very often you can apply this procedure:
Integrate by parts twice!
The original integral will appear somewhere on the right hand side.
Move it back to the left (that's when you get 5/4 circled in red) and the right hand side is free of integrals.
The part in green is the second integration by parts of the integral just to the left of the expression circled in green.

Symbolically, what it was trying to do was:
First integration by parts, (I=integral, E=expression)
I=E1+I1
Second integration by parts,
I1=E2+k.I2
But 'by chance', I2 is the same as I, so the whole thing becomes
I=E1+E2+k.I,
or
I=(E1+E2)/(1+k)

Answer 3

I think I can help you if you ask more specifically. I do know that as far as "what do I substitute" will always be the question asked. There's a rule to help you with that. It's something like ILATE Inverse Trigonometric Functions Logarithmic Functions Algebraic Functions Trigonometric Functions Exponential Functions The closer to the left, the more likely it should be u. The closer to the right, the more likely it should be dv.

Answer 4

The right-hand side has a term
-(1/4)int((e^2x)*sin x) when you expand,

and when you get this term back to the left side it adds to the 1*int(etc) which is already there.

This technique crops up quite often in integrals involving sine and/or cosine.

Answer 5

This is an alternative solution to the problem
int ( sin(x) * e^(2x) ) dx.

int ( sin(x) * e^(2x) ) dx = int(U) dV

where U = e^(2x) and dV = sin(x) dx.
Thus, dU = 2e^(2x) dx and V = -cos(x).

So, we have

int (U) dV = UV - int (V) dU
= - e^(2x) cos(x) - int ( -cos(x) * 2e^(2x) ) dx
= - e^(2x) cos(x) + 2 int ( cos(x) * e^(2x) ) dx. [eqn: 1]

Consider int ( cos(x) * e^(2x) ) dx. Then,

int ( cos(x) * e^(2x) ) dx = int (S) dT

where S = e^(2x) and dT = cos(x) dx.
Thus, dS = 2e^(2x) dx and T = sin(x).

So, we have

int (S) dT = ST - int (T) dS
= e^(2x) sin(x) - int ( sin(x) * 2e^(2x) ) dx
= e^(2x) sin(x) - 2 int ( sin(x) * e^(2x) ) dx. [eqn: 2]

If we plug [eqn: 2] to [eqn: 1], we get

int ( sin(x) * e^(2x) ) dx = - e^(2x) cos(x) + 2 [e^(2x) sin(x) - 2 int ( sin(x) * e^(2x) ) dx]
= - e^(2x) cos(x) + 2 e^(2x) sin(x) - 4 int ( sin(x) * e^(2x) ) dx].

Transposing 4 int ( sin(x) * e^(2x) ) dx to the left hand side yields to
5 int ( sin(x) * e^(2x) ) dx = - e^(2x) cos(x) + 2e^(2x) sin(x).

Dividing both sides by 5, we get
int ( sin(x) * e^(2x) ) dx = -(1/5)e^(2x) cos(x) + (2/5)e^(2x) sin(x)
= (-1/5) * ( cos(x) - 2 sin(x) ) * e^(2x).