Integrate: ∫ 8^x cot^2(8x)dx

Answer 1

∫8x cot²(8x) dx

Perform a w-substitution. Let w = 8x, then dw = 8dx

(1/8)∫w cot²(w) dw

Now, perform Integration by Parts

∫ u dv = uv - ∫v du

u = w dv = cot²(w) dw

du = dw v = ∫cot²(w) dw = -w - cotw

Therefore,

∫w cot²(w) dw = w(-w - cotw) - ∫(-w - cotw)dw

∫w cot²(w) dw = -w² - wcotw + ∫(w + cotw)dw

∫w cot²(w) dw = -w² - wcotw + (1/2)w² + ln(|sinw|) + C, where C is an arbitrary constant

∫w cot²(w) dw = -(1/2)w² - wcotw + ln(|sinw|) + C

(1/8)∫w cot²(w) dw = -(1/16)w² - (1/8)wcotw + (1/8)ln(|sinw|) + C

Recall that w = 8x

∫8x cot²(8x) dx = -(1/16)(8x)² - (1/8)(8x)cot(8x) + (1/8)ln(|sin(8x)|) + C

∫8x cot²(8x) dx = -4x² - xcot(8x) + (1/8)ln(|sin(8x)|) + C <--- ANSWER

We can check that this is the correct answer by taking the derivative and showing that we get the original integrand.

(d/dx)[-4x² - xcot(8x) + (1/8)ln(|sin(8x)|) + C] = -8x -cot(8x) +8xcsc²(8x) + (1/8)*8cos(8x)/sin(8x)

= -8x -cot(8x) + 8xcsc²(8x) + cot(8x)

The second and fourth terms cancel

= -8x + 8xcsc²(8x) = -8x + 8x[1 + cot2(8x)] using the identity csc²u = 1 + cot²u

= -8x + 8x + 8xcot2(8x) = 8xcot²(8x), which is just the original integrand.

Answer 2

No, you must integrate this function numerically.

Please provide limits of integration and then we can do Simpson's rule to approximate