ʃ (4sin²x cos²×/sin2x cos2x)dx
PLEASE HELP!!
Your expression can be simplified first.
4sin^2 x cos^2 x /(sin 2x cos 2x)
= 4 sin^2 x cos^2 x/(2sinxcosx cos2x0
= 2sinxcosx/cos2x
= sin 2x/cos 2x
so ʃ (4sin²x cos²×/sin2x cos2x)dx
= ʃ sin2x / cos2x dx
= -(1/2) ln (cos2x) + c
check:
d( -(1/2) ln (cos2x)) /dx
= (-1/2) (1/cos2x)(-2sin2x)
= sin2x/cos2x as needed
To integrate the given expression, let's simplify the integrand first.
Starting with the numerator, we have 4sin²(x)cos²(x).
Utilizing the identity that sin²(x) + cos²(x) = 1, we can rewrite sin²(x) = 1 - cos²(x).
So, the numerator becomes:
4(1 - cos²(x))cos²(x)
Expanding the equation, we get:
4cos²(x) - 4cos⁴(x)
Now, let's simplify the denominator, sin(2x)cos(2x).
Using the double-angle identity for sine, sin(2x) = 2sin(x)cos(x).
Therefore, the denominator becomes:
2sin(x)cos(x) * 2cos(x)sin(x) = 4sin(x)²cos(x)²
Now, let's substitute the simplified numerator and denominator back into the original integral expression:
∫ (4cos²(x) - 4cos⁴(x)) / (4sin(x)²cos(x)²) dx
Simplifying further, we can cancel out the 4's:
∫ (cos²(x) - cos⁴(x)) / (sin²(x)cos²(x)) dx
Now, we can split the expression into two separate integrals:
∫ cos²(x) / (sin²(x)cos²(x)) dx - ∫ cos⁴(x) / (sin²(x)cos²(x)) dx
Simplifying the integrals, we have:
∫ dx / sin²(x) - ∫ cos²(x) / sin²(x) dx
Using the trigonometric identity csc²(x) = 1/sin²(x), we can rewrite the first integral as:
∫csc²(x) dx - ∫ cos²(x) / sin²(x) dx
Now, integrating the first term gives:
- cot(x) - ∫ cos²(x) / sin²(x) dx
To integrate the second term, we can use the identity cos²(x) = 1 - sin²(x):
- cot(x) - ∫ (1 - sin²(x)) / sin²(x) dx
Simplifying, we have:
- cot(x) - ∫ (1/sin²(x) - sin²(x)/sin²(x)) dx
Which further simplifies to:
- cot(x) - ∫ csc²(x) dx + ∫ dx
Integrating the first term gives -cot(x), and integrating dx gives x:
- cot(x) + -csc(x) + x + C
Therefore, the final answer is:
-cot(x) - csc(x) + x + C
(Note: C is the constant of integration)