What is the indefinite integral of:
(12sin(3x)cscx)dx
If I could have two things given to me please, I'd be grateful. The two things are:
1) Trick/Technique to Use
2) First 2 or 3 steps
Thanks in advanced.
Express sin(3x) in terms of sin(x).
exp(ix) = cos(x) + i sin(x) (1)
Take the third power of both sides:
exp(3ix) = [cos(x) + i sin(x)]^3
From (1) it follows that the imaginary part of the left hand side is sin(3x)
Expand the right hand side and take the imaginary part. You then find:
sin(3x) = -sin^3(x) + 3sin(x)cos^2(x)
= 3 sin(x) - 4 sin^3(x)
It then follws that:
sin(3x)/sin(x) =
3 - 4 sin^2(x)
which is almost trivial to integrate.
Out of curiosity, where did the 'i' come from, and what technique did you use?
Also, how did you write sin(3x) In terms of sin(x)?
To find the indefinite integral of (12sin(3x)cscx)dx, you can use the technique of u-substitution.
Before we dive into the steps, it's important to note that csc(x) is the reciprocal of sin(x). Therefore, we can rewrite the given expression as:
(12sin(3x) * 1/sin(x))dx
Now, let's go through the first few steps:
Step 1: Identify the substitution
Let's choose u = sin(x). This allows us to rewrite the integral as:
(12sin(3x) * 1/u)du
Step 2: Find the differential
To do this, we need to differentiate both sides of the equation u = sin(x) with respect to x.
du/dx = cos(x)
Step 3: Solve for dx
Rearrange the equation to solve for dx.
dx = du/cos(x)
Step 4: Substitute and rewrite the integral
Using the substitutions in steps 1, 2, and 3, we can rewrite the original integral as:
∫ (12sin(3x) * 1/u) * (du/cos(x))
Step 5: Simplify
Now, we can cancel out sin(x) and cos(x) terms, resulting in:
∫ (12sin(3x) * 1/u) * (du/(1/u))
Simplifying further gives:
∫ 12sin(3x) * u du
Step 6: Integrate
At this point, we can integrate the expression with respect to u:
12 ∫ sin(3x) u du
After integrating, we can finally substitute back u = sin(x) to find the complete answer.
Note: The remaining steps involve evaluating the integral, which may require additional techniques.