Integrate Sin(3x)Cos(4x) - No Trig Identities Needed! Going in a Loop!

In this video, we use the traditional method of integration by parts to find the antiderivative of a product of two different trigonometric function: sin(3x)cos(4x).

It is presented in textbooks that we are using the product-to-sum formula to turn this expression into a sum. But actually, we don't need to.

It does not matter whether we use u = sin(3x) or u = cos(4x), we are still going to get back to the original problem. So it looks like we are going in a loop!

See another video in which we go in a loop when integrating e^(-2x)sin(3x):

See how to integrate this without using integration by parts:

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[davidbrisbane]

sin(ɑ)cos(β) =½[sin(ɑ + β) + sin(ɑ - β)]

So, sin(3x)cos(4x)
= ½[sin(7x) + sin(-x)]
= ½[sin(7x) - sin(x)]

So, ∫sin(3x)cos(4x) dx
= ½∫[sin(7x) - sin(x)] dx
= - (1/14)*cos(7x) + (1/2)*cos(x)
= (1/2)*cos(x) - (1/14)*cos(7x)

[carultch]

Generalizing this method for this kind of integral gives interesting results. Consider integral sin(a*x)*cos(b*x) dx.

When frequencies of sine & cosine are different, going through this method, you get:
[b*sin(a*x)*sin(b*x) + a*cos(a*x)*cos(b*x)]/(b^2 - a^2) + C

When both functions have the same frequency, you'll get a divide by zero error in the above formula. However, you can still use a looper to integrate a sine & cosine product of the same frequency. Stop one row sooner, and your division by zero error goes away.

S ____ D I
+ __ sin(a*x) __ cos(a*x)
- ____ a*cos(x) __ 1/a*sin(x)

Spot the original integral, call it I, and solve for I:
I = 1/a*sin(a*x)^2 - I
2*I = 1/a*sin(a*x)^2
I = 1/(2*a)*sin(x)^2

Add +C and we have our result:
1/(2*a)*sin(x)^2 + C

[lovelyshimly]

Can I do the same order for my second one like my first one