Calculus II - Trigonometric Integrals

[GreenPrint]

Homework Statement

Evaluate integral( sin^3(x) cos^5(x) ) dx

Homework Equations

sin^2(x) + cos^2(x) = 1
integral x^n dx = x^(n+1)/(n+1) + c
d/dx cos(x) = -sin(x)
a^n*a^m=a^(n+m)

The Attempt at a Solution

I got -cos^6(x)/6+cos^8(x)/8+c
Apparently I did something wrong
SEE ATTACHMENT

Thank you for any assistance! I really don't see what I'm doing wrong. I believe that the method I chose to evaluate this integral is overly simple but mathematically correct and I don't know what I'm doing wrong.

 

[Ray Vickson]

Homework Statement

Evaluate integral( sin^3(x) cos^5(x) ) dx

Homework Equations

sin^2(x) + cos^2(x) = 1
integral x^n dx = x^(n+1)/(n+1) + c
d/dx cos(x) = -sin(x)
a^n*a^m=a^(n+m)

The Attempt at a Solution

I got -cos^6(x)/6+cos^8(x)/8+c
Apparently I did something wrong
SEE ATTACHMENT

Thank you for any assistance! I really don't see what I'm doing wrong. I believe that the method I chose to evaluate this integral is overly simple but mathematically correct and I don't know what I'm doing wrong.

Your answer is correct; why do you think it is wrong? Here it is in Maple 9.5:
Int( sin(x)^3 *cos(x)^5,x);J1:=value(%);
/
|
| sin(x)^3 cos(x)^5 dx
|
/
J1 := -1/8 sin(x)^2 cos(x)^6 - 1/24 cos(x)^6 <---indef. integral
simplify(diff(J1,x));

sin(x)^3 cos(x)^5 <---- derivative is OK
expand(subs(sin(x)^2=1-cos(x)^2,J1));

-1/6 cos(x)^6 + 1/8 cos(x)^8
That's your answer.

RGV

 

[tiny-tim]

Hi GreenPrint!

(try using the X² icon just above the Reply box )

Looks ok to me.

(what's the official answer?)

 

[GreenPrint]

Ahhh wolfram alpha is giving me answers that don't match mine and so I think that I am wrong...

wolfram alpha
http://www.wolframalpha.com/input/?i=integral%28+sin^3%28x%29+cos^5%28x%29+%29+dx
(-72cos(2x)-12cos(4x)+8cos(6x)+3cos(8x))/3072 + constant

and my answer
-cos^6(x)/6+cos^8(x)/8+c

they don't match which is fine but they don't appear to be equivalent to each other

lol i should start using that button, thanks

 

[tiny-tim]

Hi GreenPrint!

(-72cos(2x)-12cos(4x)+8cos(6x)+3cos(8x))/3072 + constant

and my answer
-cos^6(x)/6+cos^8(x)/8+c

They're probably the same …

try rewriting the first one using 1 + cos(2x) = 2cos²x etc …

what do you get? ​

 

[GreenPrint]

-cos^6(5)/6+cos^8(5)/8 is about -8.15876413*10^-5
(-72cos(2*5)-12cos(4*5)+8cos(6*5)+3cos(8*5))/3072 is about .0178220582

umm hmmm i don't think i have seen that formula in over 3 years... uh let's see

I don't know I don't think I remember enough of the pre calc formulas to do that >_>