Integrating Sin^2x Secx: Best Method

[robierob12]

In a diff. Equation I am doing I have to integrate

<br /> <br /> \sin ^2 x\sec x<br /> <br /> <br />


which I can remember how to do.

What is the best method for this?

 

[Moridin]

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

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

 

[bob1182006]

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

 

[robierob12]

<br /> <br /> \int {\frac{{\sin ^2 x}}{{\cos x}}} dx<br /> <br />

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

 

[robierob12]

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

 

[rock.freak667]

Try using sin^2x=1-cos^2x and then look for the integral of secx

 

[robierob12]

....

 

[robierob12]

Try using sin^2x=1-cos^2x and then look for the integral of secx

not sure what your getting at there

<br /> \[<br /> \int {\frac{{\sin ^2 x}}{{\cos x}}} dx = \int {\frac{{1 - \cos ^2 x}}{{\cos x}}dx = ?} <br /> \]<br />

 

[robierob12]

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.

 

[bob1182006]

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.

 

[robierob12]

\[<br /> \begin{array}{l}<br /> \int {\sin ^2 x} \sec xdx = \int {(1 - \cos ^2 x)\sec xdx = \int {\sec x - \sec x\cos x\cos xdx} } \\ <br /> = \int {\sec x - \cos xdx} \\ <br /> \end{array}<br /> \]<br />

Does this work?

 

[bob1182006]

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

 

[Hells_Kitchen]

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

 

[robierob12]

Thanks guys.