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Integration Help

  • Thread starter robierob12
  • Start date
  • #1
48
0
In a diff. Equation im doing I have to integrate

[tex]

\sin ^2 x\sec x


[/tex]


which I can remember how to do.

What is the best method for this?
 

Answers and Replies

  • #2
664
3
Sec(x) = 1/cos x => (sin x)^2 / cos x

Notice how one of them is the derivative of the other.
 
  • #3
492
1
with respect to x I'm guessing?

[tex]\int sin^2 x sec x dx=\int \frac{sin^2 x}{cos x} dx[/tex]

just do a substitution. you probably didn't see the substitution right away with the secant, best to change to sin and cos first
 
  • #4
48
0
[tex]

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

[/tex]

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
 
  • #5
48
0
or if let u=cosx im left with an exta sinx after substitution...
 
  • #6
rock.freak667
Homework Helper
6,230
31
Try using sin^2x=1-cos^2x and then look for the integral of secx
 
  • #7
48
0
.............
 
  • #8
48
0
Try using sin^2x=1-cos^2x and then look for the integral of secx
not sure what your getting at there

[tex]
\[
\int {\frac{{\sin ^2 x}}{{\cos x}}} dx = \int {\frac{{1 - \cos ^2 x}}{{\cos x}}dx = ?}
\]
[/tex]
 
  • #9
48
0
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.
 
  • #10
492
1
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.
 
  • #11
48
0
[tex]\[
\begin{array}{l}
\int {\sin ^2 x} \sec xdx = \int {(1 - \cos ^2 x)\sec xdx = \int {\sec x - \sec x\cos x\cos xdx} } \\
= \int {\sec x - \cos xdx} \\
\end{array}
\]
[/tex]

Does this work?
 
  • #12
492
1
yes, now you'd need to find the integral for sec x
 
  • #13
which i think its just the ln abs(secX + tanX)
 
  • #14
48
0
Thanks guys.
 

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