see attached

Printable View

- Mar 18th 2006, 12:02 PMfrozenflamesHelp In Calculus Needed
see attached

- Mar 18th 2006, 10:39 PMearbothQuote:

Originally Posted by**frozenflames**

you have to find $\displaystyle \int {sin(x) \cdot cos(y)\ dx$

Substitute $\displaystyle u=sin(x)\ \Longrightarrow \ \frac{du}{dx}=cos(x)\ \Longrightarrow \ du=cos(x) \cdot dx$

The integral is now $\displaystyle \int {u \cdot du}=\frac{1}{2}(u)^2+c$

Re-substitute and you'll get: $\displaystyle \int {sin(x) \cdot cos(y)\ dx=\frac{1}{2}(sin(x))^2+c$

So it is answer A.

Greetings

EB