- #1

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[tex]\int {\sin x \cdot \cos x{\rm{ dx}}} [/tex]

I've tried to use integration by parts but that doesn't seem to get me anywhere...any tips?

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- #1

- 424

- 0

[tex]\int {\sin x \cdot \cos x{\rm{ dx}}} [/tex]

I've tried to use integration by parts but that doesn't seem to get me anywhere...any tips?

- #2

Staff Emeritus

Science Advisor

Gold Member

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Seems to work for me... can you show your work?I've tried to use integration by parts but that doesn't seem to get me anywhere

Incidentally, there are

(And both of these ways are almost certainly shown in your book)

- #3

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I define the following:

[tex]

\[

\begin{array}{*{20}c}

{u' = - \sin x} & {v = \cos x} \\

{u = \cos x} & {v' =- \sin x} \\

\end{array}

\]

[/tex]

Then I put this into the formula:

[tex]

\[

\begin{array}{l}

\int {\sin x \cdot \cos xdx{\rm{ }} = uv - \int {uv'dx} } \\

= \cos x \cdot \cos x - \int {\cos x \cdot ( - \sin x)dx} \\

= \cos ^2 x - \int {\cos x \cdot ( - \sin x)dx} \\

\end{array}

\]

[/tex]

This integration at the end is no easier than what I started with...so what's wrong here?

[tex]

\[

\begin{array}{*{20}c}

{u' = - \sin x} & {v = \cos x} \\

{u = \cos x} & {v' =- \sin x} \\

\end{array}

\]

[/tex]

Then I put this into the formula:

[tex]

\[

\begin{array}{l}

\int {\sin x \cdot \cos xdx{\rm{ }} = uv - \int {uv'dx} } \\

= \cos x \cdot \cos x - \int {\cos x \cdot ( - \sin x)dx} \\

= \cos ^2 x - \int {\cos x \cdot ( - \sin x)dx} \\

\end{array}

\]

[/tex]

This integration at the end is no easier than what I started with...so what's wrong here?

Last edited:

- #4

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Your v' is wrong.

- #5

Gold Member

MHB

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How about U-substitution? What can you substitute for u? And **du**? (major hint)

- #6

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- #7

Gold Member

MHB

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[tex]\int{udv}=uv-\int{vdu}[/tex]

It's minus the integral of v times

- #8

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Ah, I got it now. Thanks!

- #9

Gold Member

MHB

- 4,539

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No prob. :)

- #10

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- #11

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Try: u = sinx du = cosxdx

or try the fact that sin2x = 2sinxcosx

or try the fact that sin2x = 2sinxcosx

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