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Find the integral Please HELP

  • Thread starter BuBbLeS01
  • Start date
602
0
Find the integral...Please HELP!!!!

1. Homework Statement

Find the integral involving sin, cos, sec, and tan:
A. sin^3x * sqrt(cosx)dx
B. sec^3(2x) * tan(2x)dx

2. Homework Equations



3. The Attempt at a Solution
I will start with part A first...
A. sin^3x * sqrt(cosx) dx
sin^2*x * sqrt(cosx) * sinx dx
(1-cos^2*x) * sqrt(cosx) * sinx dx
sinx * sqrt(cosx) - sinx * cosx dx

but not I don't know what to do??
 

Answers and Replies

Gib Z
Homework Helper
3,344
4
You did well, but just stop on the second last line, make a substitution u= sin x, then you have a simple integral in powers of u no?
 
602
0
So I stop here....

(1-cos^2*x) * sqrt(cosx) * sinx dx

???

I am not understanding how to use u-substitution with this? u = sinx, du = cosx but thats not it?
 
Gib Z
Homework Helper
3,344
4
SOrry sorry! I meant u=cos x, du = -sin x dx.
 
602
0
Umm but I am still not understanding lol im sorry...

(1-cos^2*x) * sqrt(cosx) * sinx dx

u=cos x, du = -sin x dx

So if I do u^1/2 the integral is 2/3 * u^3/2

but that only gets me....... sqrt(cosx) * -sinx dx

right??
 
Gib Z
Homework Helper
3,344
4
When you make the substitution,

[tex]\int (1- \cos^2 x) \sqrt{\cos x} \sin x dx = - \int (1-u^2) \sqrt{u} du = -\int ( u^{1/2} - u^{5/2}) du [/tex]

You should be able to finish it off with the power rule, and replace back in u=cos x at the end.
 
Last edited:
602
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so if I leave it at...

1 - u^2 * sqrt(u) du

I could write it as


x - cosx^2 * sqrt(cosx)
 
Gib Z
Homework Helper
3,344
4
Did you follow my last post :( ? Do you know how to integrate those terms I had in the last post?
 
602
0
woops that wasn't done right...

(x - 1/3*cosx^3) * 2/3*sqrt(cosx)^3/2

can I write it like this...I know its not completely reduced yet but I will do that.
 
Gib Z
Homework Helper
3,344
4
Well when I evaluate the integral I posted, I get [tex]2\cos^{3/2} x \left( \frac{1}{3} - \frac{\cos^2 x}{7}} \right) [/tex]
 
602
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Why am I not understanding this....

(1-u^2) * 2/3 u^3/2

so I plug u=cosx back in and integrate...

(x - 1/3*cosx^3) * 2/3*(cosx)^3/2

can you show me how you got your answer?
 
Gib Z
Homework Helper
3,344
4
Do you follow post 6? I get my answer from post 6 by directly using the power rule.
 
602
0
I thought it would be....

u^3/2 - u^5/2
 
1,750
1
[tex]\int\sin^3 x\sqrt{\cos x}dx[/tex]

[tex]\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx[/tex]

[tex]u=\cos x[/tex]
[tex]du=-\sin xdx[/tex]

[tex]-\int(1-u^2)\sqrt udu[/tex]

[tex]-\int(u^{\frac 1 2}-u^{\frac 5 2})du[/tex]
 
Gib Z
Homework Helper
3,344
4
I thought it would be....

u^3/2 - u^5/2
Check the power rule again.

[tex]\int\sin^3 x\sqrt{\cos x}dx[/tex]

[tex]\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx[/tex]

[tex]u=\cos x[/tex]
[tex]du=-\sin xdx[/tex]

[tex]-\int(1-u^2)\sqrt udu[/tex]

[tex]-\int(u^{\frac 1 2}-u^{\frac 5 2})du[/tex]
I believe that was already well established in this thread.
 
602
0
Okay I don't know why I am not seeing what you are doing but when you have...
(1-u^2) * sqrt u

don't you multiply u^1/2 by (1-u^2) ???

and get u^1/2 - u^3/2

so I don't understand why the second term is ^5/2?
 
1,750
1
BuBbLeS ... what is 1.5 and 2.5 in fractions?
 
458
0
Okay I don't know why I am not seeing what you are doing but when you have...
(1-u^2) * sqrt u

don't you multiply u^1/2 by (1-u^2) ???

and get u^1/2 - u^3/2

so I don't understand why the second term is ^5/2?
[tex]x^a\cdot x^b = x^a^+^b[/tex]

Now add 1/2 and 2.
 
602
0
OMG I feel like such an IDIOT right now...thanks!
 

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