Integral of cos^3 x?

Question

Evaluate the integral from 0 to pi/2 of cos^3 (x)dx

I got to the end and my answer was negative.

Please show step by step work.

Answer 1

[ stands for integral
[cos^3 x dx
[cos^2 x(cos(x) dx
[(1- sin^2 x)cosx dx
[cos(x)dx - [sin^2(x)cos(x)
sin(x) - [sin^2(x)d(sinx) (cos x is differential of sinx)
sin(x) - sin^3(x)/3
(sin(x) - 1/3(sin^3(x))
(sin(pi/2) - sin(0) - 1/3((sin^3(pi/2) - sin^3(0))
(1 - 0 )- 1/3((1) - 0)
1 - 1/3 = 2/3

Answer 2

You know that: cos^3 x = (cos x)(cos^2 x) and that from cos^2 x + sin^2 x = 1, we have cos^2 x = 1 - sin^2 x

We will have: cos^3 x = (cos x)(1 - sin^2 x) = cos x - (sin^2 x)(cos x)
When you apply the integration, you will get:
Integral of [cos x] = sinx and integral of [(sin^2 x)(cos x)] = (1/3)(sin^3 x)

Now you evaluate [sin x - (1/3)(sin^3 x)] from 0 to pi/2, and will get [1 - 1/3 - 0 + 0]
The final answer will be: 2/3

Answer 3

Integral Of Cos 3x

Answer 4

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I assmume that is sin(x)^2 cos(x)^3 and not sin(2x)cos(3x)dx: First break it up: sin(x)^2 cos(x)^2 cos(x) dx Pythagorean Identity for cosine: sin(x)^2 (1 - sin(x)^2) cos(x) dx Simplify: (sin(x)^2 - sin(x)^4) cos(x) dx u-substitution of u = sin(x), thus du = cos(x) dx: (u^2 - u^4) du Edit: Just in case.... (since I hate when a solutions manual skips steps) Anti-differentiate: 1/3 u^3 - 1/5 u^5 + C Then back-substitute your x values since this is an indefinite integral. /Edit answer: 1/3 sin(x)^3 - 1/5 sin(x)^5 + C

Answer 5

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RE:
Integral of cos^3 x?
Evaluate the integral from 0 to pi/2 of cos^3 (x)dx

I got to the end and my answer was negative.

Please show step by step work.

Answer 6

∫ cos^3(x) dx
= ∫ cos^2(x)*cos(x) dx
= ∫ (1-sin^2(x))cos(x) dx
= ∫cos(x) dx - ∫cos(x)sin^2(x) dx
= sin(x) - ∫cos(x)sin^2(x) dx
Let u = sin(x) ; du = cos(x) dx
∫ = sin(x) - ∫ u^2 du
= sin(x) - u^3 / 3 + c
= sin(x) - sin^3(x) / 3 + c

At π/2, this evauates to 2/3 + c
At 0 this evaluates to 0 + c

The answer is 2/3 + c - (0+c) = 2/3

Answer 7

∫cos³ x dx = (9 sin x + sin (3x))/12 + C
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