How do I find the integral of sin^3 (t) cos(tà

[courtrigrad]

How would you do: \int \sin^{3} t cos t? Would u = t\cos t?

Thanks

 

[dextercioby]

Isn't your integral just
\int \sin^{3}t \ \cos t \ dt

...?

Daniel.

 

[courtrigrad]

yes. so would u = t, dt = du [/tex]<br /> <br /> Also for \int (x-1)e^{x^{2} - 2x} woud u = x^{2} - 2x? Because I know that the answer is \frac{1}{2}(x-1)e^{x^{2} - 2x}.<br /> <br /> Thanks

 

[dextercioby]

For the first part,that's not a valid change of variable.

For the second,it's okay.

Daniel.

 

[courtrigrad]

would i apply the identity sin^{2} x = 1 - cos^{2} x? Also for the second one, once i make the substitution how do we get the \frac{1}{2} in the front?

Thanks

 

[dextercioby]

\int sin^{3}x \cos x \ dx=\int \sin^{3}x d(\sin x) =\frac{\sin^{4}x}{4} +C

As for the second,the 1/2 comes from the cancelation of the 2 which would result from the derivative of the exponent.

Daniel.

 

[courtrigrad]

ok thanks a lot. So you substituted d(sin x) = cos x so we get our elementary integrations.

Thanks