Evaluate the integral: sin^2(5x)cos^2(5x)dx
Better late than never.
I have a pile of scrap paper here with many tries at this, mostly trying to use integration by parts.
what a mess!
Then I tried this:
(sin(5x)cos(5x))^2 = (1/4)[sin(10x)]^2
we know cos 2A = 1 - 2sin^2 A
so cos20x - 1 - 2sin^2 (10x) , then
2sin^2 (10x) = 1 - cos (20x)
(1/4)[sin^2 (10x)] = 1/8 - (1/8)cos (20x)
then [integral](sin(5x)cos(5x))^2 dx
= [integral](1/8 - (1/8)cos (20x)) dx
= x/8 - (1/160)sin(20x)
To evaluate the integral ∫sin^2(5x)cos^2(5x)dx, you can use trigonometric identities to simplify the expression and then apply basic integration techniques.
Step 1: Applying Double-Angle Identity
Start by using the double-angle identity for sine and cosine:
sin^2(θ) = (1 - cos(2θ))/2
cos^2(θ) = (1 + cos(2θ))/2
Applying these identities to the integral, we get:
∫(1 - cos(10x))/2 * (1 + cos(10x))/2 dx
Simplifying further:
∫(1 - cos^2(10x))/4 dx
= ∫(1/4 - cos^2(10x)/4) dx
= 1/4 ∫(1 - cos^2(10x)) dx
Step 2: Applying Power-Reducing Identity
Next, apply the power-reducing identity for cosine:
cos^2(θ) = (1 + cos(2θ))/2
Substituting this identity into the integral, we have:
1/4 ∫(1 - (1 + cos(20x))/2) dx
Simplifying further:
1/4 ∫(1 - 1/2 - cos(20x)/2) dx
= 1/4 ∫(1/2 - cos(20x)/2) dx
Step 3: Integrating the Expression
Now, we can integrate the expression:
1/4 ∫(1/2 - cos(20x)/2) dx
= 1/4 [(1/2)x - (1/2)(1/20)sin(20x)] + C
= 1/4 [(x/2) - (1/40)sin(20x)] + C
Final Answer:
Therefore, the integral ∫sin^2(5x)cos^2(5x)dx evaluates to (x/8) - (1/160)sin(10x) + C, where C is the constant of integration.