# Integration Help

In a diff. Equation im doing I have to integrate

$$\sin ^2 x\sec x$$

which I can remember how to do.

What is the best method for this?

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Sec(x) = 1/cos x => (sin x)^2 / cos x

Notice how one of them is the derivative of the other.

with respect to x I'm guessing?

$$\int sin^2 x sec x dx=\int \frac{sin^2 x}{cos x} dx$$

just do a substitution. you probably didn't see the substitution right away with the secant, best to change to sin and cos first

$$\int {\frac{{\sin ^2 x}}{{\cos x}}} dx$$

let u=sinx du=coxdx

I don't see this working becuase i have (1/cosx)dx not cosx dx....

Am I looking at the wrong substitution?

Rob

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

rock.freak667
Homework Helper
Try using sin^2x=1-cos^2x and then look for the integral of secx

.............

Try using sin^2x=1-cos^2x and then look for the integral of secx
not sure what your getting at there

$$$\int {\frac{{\sin ^2 x}}{{\cos x}}} dx = \int {\frac{{1 - \cos ^2 x}}{{\cos x}}dx = ?}$$$

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.

hm..didn't notice that sin^2 x....

what rock.freak.667 meant was to split up that fraction and you'll have integral of sec x -cos x. so you'd have to find the integral of sec x.

$$$\begin{array}{l} \int {\sin ^2 x} \sec xdx = \int {(1 - \cos ^2 x)\sec xdx = \int {\sec x - \sec x\cos x\cos xdx} } \\ = \int {\sec x - \cos xdx} \\ \end{array}$$$

Does this work?

yes, now you'd need to find the integral for sec x

which i think its just the ln abs(secX + tanX)

Thanks guys.