Question

Question #e1ae5

Answer 1

`int_0^(pi/4)16sin^4(x)cos^2(x)dx=pi/4-1/3`

Explanation:

First, let's find the indefinite integral `int16sin^4(x)cos^2(x)dx`. To do so, we will make use of the formulas

• `sin^2(x) = (1-cos(2x))/2`
• `sin(2x) = 2sin(x)cos(x)`

`int16sin^4(x)cos^2(x)dx = 16intsin^2(x)(sin(x)cos(x))^2dx`

`=16intsin^2(x)(1/2sin(2x))^2dx`

`=4intsin^2(x)sin^2(2x)dx`

`=4int(1-cos(2x))/2sin^2(2x)dx`

`=2int(sin^2(2x)-sin^2(2x)cos(2x))dx`

`=2intsin^2(2x)dx - 2intsin^2(2x)cos(2x)dx`

Let's evaluate these integrals separately.

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`2intsin^2(2x)dx = 2int(1-cos(4x))/2dx`

`=intdx - intcos(4x)dx`

`=x-sin(4x)/4+C`

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For the second integral, we make the substitution

`u = sin(2x) => du = 2cos(2x)dx`

Then

`2intsin^2(2x)cos(2x)dx = intu^2du`

`=u^3/3+C`

`=sin^3(2x)/3+C`

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Putting the above together, we get

`int16sin^4(x)cos^2(x)dx`

`= 2intsin^2(2x)dx - 2intsin^2(2x)cos(2x)dx`

`=x-sin(4x)/4-sin^3(2x)/3+C`

We can now evaluate the definite integral.

`int_0^(pi/4)16sin^4(x)cos^2(x)dx = [x-sin(4x)/4-sin^3(2x)/3]_0^(pi/4)`

`=(pi/4-sin(pi)/4-sin^3(pi/2)/3)`

`-(0-sin(0)/4-sin^3(0)/3)`

`=pi/4-0/4-1/3-0+0+0`

`=pi/4-1/3`