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1. Evaluate the integral of (sin(8x))^3*(cos(8x))^2*dx. Use C to denote an arbitrary constant.

2. Evaluate the integral of (cos(5x))^3*dx. Use C to denote an arbitrary constant.

3. Evaluate the integral of csc(3x)dx. Use C to denote an arbitrary constant.

4. Evaluate the integral of (2-2sinx)/cosx dx. Use C to denote an arbitrary constant.

2007-05-30 16:18:07 · 3 answers · asked by aSnxbByx113 2 in Science & Mathematics Mathematics

3 answers

These things are really tedious. No one's going to do all your HW like that. But I'll give you some tips.

Look for multiples of cos and sin rather than powers.
eg if you have cos^2 (x). It's best to use one of the trig identities go rewrite it, such as
cos^2 (x) = 1/2 + 1/2*cos(2x).
That's easier to integrate.

2007-05-30 16:45:57 · answer #1 · answered by Dr D 7 · 1 0

Integral ( sin^3(8x) cos^2(8x) dx )

To integrate this, first break off a sin(8x) and then use substitution.

Integral ( sin^2(8x) cos^2(8x) sin(8x) dx )

Use the identity sin^2(8x) = 1 - cos^2(8x)

Integral ( [1 - cos^2(8x)] cos^2(8x) sin(8x) dx )

Let u = cos(8x). Then
du = -8sin(8x) dx, so
(-1/8) du = sin(8x) dx

Integral ( [1 - u^2] [u^2] (-1/8) du )

Factor out (-1/8).

(-1/8) Integral ( [1 - u^2] [u^2] du )

Expand.

(-1/8) Integral ( (u^2 - u^4) du )

Use the reverse power rule.

(-1/8) ( (1/3)u^3 - (1/5)u^5 ) + C

Distribute the -1/8.

(-1/24)u^3 + (1/40)u^5 + C

But u = cos(8x), so our final answer is

(-1/24)cos^3(8x) + (1/40)cos^5(8x) + C

2. Break off a cos(5x) and do as #1.

3. The integral of csc(x) is equal to ln|csc(x) - cot(x)| + C. Use this knowledge to find the integral of csc(3x).

4. Use identities to manipulate this into a more integrable form.

2007-05-30 23:52:36 · answer #2 · answered by Puggy 7 · 0 0

(sin(8x))^3*(cos(8x))^2*dx
Let u = 8x. Then du = 8dx --> dx = du/8
so .125sin^3 u * cos^2 u du
Now use the formula:
int[sin^mucos^nu du = - [sin^(m-1) u cos^(n+1) u]/(m+n) +
(m-1)/(m+n) *integ[sin^m-2 u cos^n u du]

(cos(5x))^3*dx
let u = 5x so du = 5dx --> dx =du/5
so .2cos^3u du = .2sin u+C = .2sin 5x +C

csc(3x)dx
Use integ csc u du = ln tan(u/2) +C

(2-2sinx)/cosx dx
=[2sec x -2tanx]dx
= 2ln(sec x+ tan x) - 2(-ln cos x) + C
ln(sec x+tanx)^2 + ln cos^2 x +C

2007-05-30 23:55:24 · answer #3 · answered by ironduke8159 7 · 0 0

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