- #1

- 48

- 0

[tex]

\sin ^2 x\sec x

[/tex]

which I can remember how to do.

What is the best method for this?

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- Thread starter robierob12
- Start date

- #1

- 48

- 0

[tex]

\sin ^2 x\sec x

[/tex]

which I can remember how to do.

What is the best method for this?

- #2

- 670

- 3

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

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

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

- #3

- 492

- 1

[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

\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

- 6,223

- 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

don't know why this one is stumping me so much.

- #10

- 492

- 1

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

\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

- 62

- 0

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

- #14

- 48

- 0

Thanks guys.

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