In this question, you will solve a definite integral problem using the method of substitution.

Question

In this question, you will solve a definite integral problem using the method of substitution.

∫ sin(4x)cos6(4x)dx from pi to pi/3

(a) Make the substitution u= cos4x], and write the integrand as a function of u. ∫ sin(4x)cos6(4x)dx =

(b) Hence solve the integral as a function of u. You do not need a constant of integration for definite integrals.

When x=π/3=, what is

(c) the value of u?

(d) ... and part(b) ?

When x=π, what is

(e) the value of u?

(f) ... and part (b) ?

(g) Hence give the value of the definite integral

I get the correct answer for (a),(b),(c),(e) and (f) but i get wrong answer for (d) and (g)

Answer 1

u = cos(4x), du = –4 sin(4x) dx which means that sin(4x) dx = –(1/4)du

(a) ∫ sin(4x)cos⁶(4x)dx = –(1/4)∫ u⁶ du

(b) –(1/4)∫ u⁶ du = –(1/4)(u⁷/7) = –(u⁷/28)

(c) x = π/3 means that u = cos(4x) = cos(4π/3) = –1/2

(d) u = –1/2 means that –(u⁷/28) = –(–1/2)⁷(1/28) = –(–1/128)(1/28) = 1/3584

(e) x = π means that u = cos(4x) = cos(4π) = 1

(f) u = 1 means that –(u⁷/28) = –(1)⁷(1/28) = –1/28

(g) ∫ sin(4x)cos⁶(4x)dx from π to π/3 = –(1/4)∫ u⁶ du from 1 to –1/2 =

= 1/3584 – (–1/28) = 1/3584 + 1/28 = 1/3584 + 128/(128×28)= 1/3584 + 128/3584 = 129/3584