Evaluate the integral using trigonometric substitution. Please show a complete solution. Thanks for your help.

Question

The integral (with upper limit pi/4 and lower limit 0) of (sinx)^4 (cosx)^2 dx.
The final answer is (3pi - 4)/192.

Answer 1

You need to use the following trig identities:

(sin x)^2 = [1-cos(2x)]/2
(cos x)^2 = [1+cos(2x)]/2

Substituting these into the integrand, you get:

([1-cos(2x)]/2)^2*
([1+cos(2x)]/2)dx

Simplifying, you get:

(1/8)(1 - cos(2x) - [cos(2x)]^2 + [cos(2x)]^3)dx

Now you have 4 separate integrals to solve:

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int[1*dx] = x

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int[cos(2x)dx] -------> let u = 2x, du = 2dx ----> dx = (1/2)du

= (1/2)*int[cos(u)du] = (1/2)sin(2x)

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int[(cos(2x))^2dx] ----> use (cos x)^2 = [1+cos(2x)]/2 to simplify:

int([1+cos2(2x)]/2*dx) = (1/2)*[int(1) + int(cos(4x)dx)]

let u = 4x, du = 4dx --> dx = (1/4)du

= (1/2)x + (1/2)(1/4)int(cos(u)du) =

(1/2)x + (1/8)sin(4x)

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int([cos(2x)]^3dx), use the same methods above, you get:

(3/8)sin(2x) + (1/24)sin(6x)

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Combining everything together, you get:

(1/16)x - (1/64)sin(2x) - (1/64)sin(4x) + (1/192)sin(6x)

Evaluating at upper limit (pi/4):
(pi/64) - (1/64)sin(pi/2) - (1/64)sin(pi) + (1/192)sin(3*pi/2)

= (pi/64) - (1/64) - (1/192) = (3pi - 4)/192

Evaluating at lower limit (0):
0 - (1/64)sin(0) - (1/64)sin(0) + (1/192)sin(0) = 0

Thus, your answer is:

(3pi - 4)/192, which is the answer your desired answer

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Hope this helps

Answer 2

(sinx)^4 (cosx)^2
[(sinx)^4 (cosx)^4]/(cosx)^2
[sinxcosx]^4/(cosx)^2
[(2sinxcosx)/2]^4/(cosx)^2
[sin2x]^4/16(cosx)^2
now sin2x=2tanx/1+(tanx)^2
so we get
[2tanx/1+(tanx)^2]^4/16(cosx)^2
{[2tanx/1+(tanx)^2]^4(secx)^2}/16
now plug tanx=t
(secx)^2dx=dt
I hope you can solve it now