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How would you begin to integrate this problem?
cos^3(x) * sqrt( sin (x) ) dx

Thanks!

2007-01-29 10:20:38 · 5 answers · asked by AlaskaGirl 4 in Science & Mathematics Mathematics

5 answers

Integral (cos^3(x) sqrt(sin(x)) dx)

Your first step would be to break off a cos(x). I'm going to insert this cos(x) at the tail end beside the dx to show you something later.

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

Now, I'm going to change cos^2(x) into 1 - sin^2(x).

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

Now, we use substitutionl

Let u = sin(x)
du = cos(x) dx {Note: this is precisely the tail end of our integer! We can replace the tail end entirely with du.}

Integral ( [1 - u^2] sqrt(u) du )

Note that sqrt(u) = u^(1/2), so

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

Distribute the u^(1/2). Note that multiplied with u^2, we have
u^(2 + 1/2), or u^(5/2).

Integral (u^(1/2) - u^(5/2)) du

Integrate normally.

(2/3) u^(3/2) - (2/7) u^(7/2) + C

And now, substitute u = sin(x) back.

(2/3) [sin(x)]^(3/2) - (2/7) [sin(x)]^(7/2) + C

2007-01-29 10:40:08 · answer #1 · answered by Puggy 7 · 0 0

Definitely a candidate for a change of variables. Take

u = sin x; so 1 - u^2 = cos^2x,

the integrand will turn into

(u^2 - 1) sqrt u du = u^(3/2) - u^(1/2) du,

which you should know how to integrate.

2007-01-29 18:40:39 · answer #2 · answered by acafrao341 5 · 0 0

Let cos^3(x) be cosx * cos^2(x)
Then convert cos^2(x) into 1-sin^2(x) [trig identity]

so you have (1-sin^2(x))*sin^1/2(x)*cosx dx
Simplify (1-sin^2(x))*sin^1/2(x):
(sin^1/2(x)-sin^5/2(x))cosx dx
Then use u substitutions:
let u = sinx
du = cosx dx
sin^1/2(x) = u^1/2
sin^5/2(x) = u^5/2

So you get the integral of:
(u^1/2 - u^5/2) du

Which becomes:
2/3 u^3/2 - 2/7 u^7/2 + c
Replacing the u's with sinx:
2/3 sin^3/2(x) = 2/7 sin^7/2(x) + c

2007-01-29 18:43:43 · answer #3 · answered by J 2 · 0 0

Integrate (cos³x)(√sin x) with respect to x.

First rearrange terms to make it easier to integrate.

(cos³x)(√sin x) = (cos x)(1 - sin²x)(√sin x)
= (cos x)[(√sin x) - (sin x)^(5/2)]

Now we can integrate.

∫(cos³x)(√sin x)dx = ∫(cos x)[(√sin x) - (sin x)^(5/2)]dx

Let
u = sin x
du = (cos x)dx

= ∫[√u - u^(5/2)]du = (2/3)u^(3/2) - (2/7)u^(7/2) + C
= (2/3)(sin x)^(3/2) - (2/7)(sin x)^(7/2) + C

2007-01-29 18:52:34 · answer #4 · answered by Northstar 7 · 0 0

idk, only in algebra......10 pts?

2007-01-29 18:49:44 · answer #5 · answered by ♥.:ily:.♥ 1 · 0 2

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