(24cos(8x))/(1+(sin^2(8x)))
Integrated from (pi/16,-pi/16
Try substitution.
Let u = sin 8x
du = 8 cos 8x dx
That makes the integral
3 du/(1+u^2)
with limits from u = 1 to -1
That should be an easy one.
Thank you. That made it so much simpler.
To integrate the given expression, we can use a substitution method. Let's break it down step by step:
1. Let's start with the given expression:
∫ (24cos(8x))/(1+(sin^2(8x))) dx
2. Notice that the denominator 1 + (sin^2(8x)) is equivalent to cos^2(8x), using the identity cos^2(x) + sin^2(x) = 1.
3. Substitute cos^2(8x) for 1 + (sin^2(8x)). The integral now becomes:
∫ (24cos(8x))/cos^2(8x) dx
4. Using the quotient rule for trigonometric functions, we can simplify further:
∫ (24cos(8x))/(cos^2(8x)) dx = ∫ 24sec(8x) dx
5. Now, we have a trigonometric function, sec(8x), in the integrand. To evaluate this integral, we can use a u-substitution.
6. Let u = 8x, then du = 8 dx. We can rewrite the integral as:
(1/8) ∫ 24sec(u) du
7. Simplify the expression:
(1/8) * 24 ∫ sec(u) du = (3/2) ∫ sec(u) du
8. The integral of sec(u) can be found using the natural logarithm function:
(3/2) ln |sec(u) + tan(u)| + C
9. Now, substitute u back in:
(3/2) ln |sec(8x) + tan(8x)| + C
10. Finally, we can evaluate the definite integral by substituting the limits of integration:
[ (3/2) ln |sec(8x) + tan(8x)| ] from -π/16 to π/16
= (3/2) ln |sec(8(π/16)) + tan(8(π/16))| - (3/2) ln |sec(8(-π/16)) + tan(8(-π/16))|
= (3/2) ln |sec(π/2) + tan(π/2)| - (3/2) ln |sec(-π/2) + tan(-π/2)|
= (3/2) ln |∞| - (3/2) ln |-∞|
Since sec(π/2) = 1/0 = ∞ and sec(-π/2) = -1/0 = -∞
= (3/2) ln |∞| - (3/2) ln |-∞|
= (3/2) ln |∞| - (3/2) ln |∞|
= (3/2) ∞ - (3/2) ∞
= ∞ - ∞
The result of this integral is undefined.