Recent content by robierob12

Thanks guys.

\[ \begin{array}{l} \int {\sin ^2 x} \sec xdx = \int {(1 - \cos ^2 x)\sec xdx = \int {\sec x - \sec x\cos x\cos xdx} } \\ = \int {\sec x - \cos xdx} \\ \end{array} \] Does this work?

I just tried integration by parts and I end up going in a circle... don't know why this one is stumping me so much.

not sure what your getting at there \[ \int {\frac{{\sin ^2 x}}{{\cos x}}} dx = \int {\frac{{1 - \cos ^2 x}}{{\cos x}}dx = ?} \]

....

or if let u=cosx I am left with an exta sinx after substitution...

\int {\frac{{\sin ^2 x}}{{\cos x}}} dx let u=sinx du=coxdx I don't see this working because i have (1/cosx)dx not cosx dx... Am I looking at the wrong substitution? Rob

In a diff. Equation I am doing I have to integrate \sin ^2 x\sec x which I can remember how to do. What is the best method for this?

it's always the obvious. how funny

Homework Statement Check to see if the (vector space) set of rational numbers is closed under addition and scalar multiplication Homework Equations The book says this holds for addition but fails for scalar multiplication. The Attempt at a Solution Im a little confused. You can...

Im doing a differential equation and end up with my last step, which is an integration. I can't seem to remember how to integrate this one int. (tanx)/(1-cosx) hmmm [(sinx)/(cosx)] / (1-cosx) u = cosx du = -sinx dx throw a negative outside the integral...

Determine the region in the xy-plane for which (1+y^3)y' = x^2 This has a unique solution. ----------------------------------------- Not really understanding what this is asking of me? I solved this as a seperable diff. Equation. (1+ y^3)dy = (x^2) dx y + (1/4)y^4 =...

Solve the Differential Equation (1-cos(x))dy + (2ysin(x) - tan(x)) dx = 0 I have tried this two ways so far and either way that I do it does not look correst. First way: As a seperable DE. (1-cos(x))dy/dx = 2ysin(x) - tan(x) simplifying I eventually get... dx - sec(x)dx =...

It's funny sometimes how easy something can end up being... just pluging in a point. Thanks, Rob

Im having a little issue figuring out this intial value problem. Solve the Initial Value Problem y' = (3x^2)/[(3y^2)-4] where y(0)=1 Looks like I can solve it as a seperable DE. dy/dx = (3x^2)/[(3y^2)-4] [(3y^2)-4] dy = (3x^2) dx Integrating both sides... (y^3) - 4y =...