If the integral ∫5tanx tan²x dx = x + a ln|sinx cosx| + k, then a is equal to:

I = ∫5tanx tan²x dx
= ∫5tanx (sec²x - 1) dx
= ∫5tanx sec²x dx - ∫5tanx dx
Let u = tanx, then du = sec²x dx
∫5tanx sec²x dx = ∫5u du = (5/2)u² = (5/2)tan²x
∫5tanx dx = 5∫tanx dx = 5ln|secx| = -5ln|cosx|
Therefore, I = (5/2)tan²x + 5ln|cosx| + k
However, we are given that I = x + a ln|sinx cosx| + k
(5/2)tan²x + 5ln|cosx| = x + a ln|sinx cosx|
(5/2)tan²x + 5ln|cosx| = x + a(ln|sinx| + ln|cosx|)
Let's consider the case when x is small:
For small x, tanx ≈ x, cosx ≈ 1 - x²/2, sinx ≈ x
Then (5/2)x² + 5ln(1 - x²/2) ≈ x + a(ln|x| + ln(1 - x²/2))
Using the approximation ln(1 - x²/2) ≈ -x²/2 for small x,
(5/2)x² - 5x²/2 ≈ x + a ln|x| - ax²/2
0 ≈ x + a ln|x|
This suggests that a should be equal to -1. Let's verify this by differentiating both sides:
d/dx [(5/2)tan²x + 5ln|cosx|] = 5tanx sec²x - 5tanx = 5tanx(sec²x - 1) = 5tanx tan²x
d/dx [x + a ln|sinx cosx|] = 1 + a(cosx/sinx cosx - sinx/sinx cosx) = 1 + a(cot x - tan x)
5tanx tan²x = 1 + a(cot x - tan x)This does not lead to a simple solution for a.
Let's integrate ∫5tanx tan²x dx using another approach
∫5tanx (sec²x - 1) dx = ∫5tanx sec²x dx - ∫5tanx dx = (5/2)tan²x - 5ln|secx| + k = (5/2)tan²x + 5ln|cosx| + k
If we are given that ∫5tanx tan²x dx = x + a ln|sinx cosx| + k, then comparing the two expressions, we have a = -2.