Help with this small integral step..

[OrangeOne]

Im solving an integral problem and a part of it means I need to solve this:

? sinxcosx dx

What is the anti-derivative of sinxcosx..I just don't get it?

Please help!

 

[Deleted member 4993]

Im solving an integral problem and a part of it means I need to solve this:

? sinxcosx dx

What is the anti-derivative of sinxcosx..I just don't get it?

Please help!

Do you know wht the anti-derivative of sin(x)?

Also remember:

$\displaystyle sin(x)\cdot cos(x) \ = \ \frac{1}{2}\cdot sin(2x)$

 

[lookagain]

? sinxcosx dx

What is the anti-derivative of sinxcosx..

Some choices:
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$\displaystyle A) \ \ 2sinxcosx \ = \ sin(2x), \ so\ sinxcosx\ = \ \frac{1}{2} sin(2x) \longrightarrow$

\(\displaystyle \int{\frac{1}{2}sin(2x) \ dx \ . \ . \ . \ . \ And \ then \ try \ that.\)

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$\displaystyle B) \ \ The \ derivative \ of \ sinx \ is \ cosx.$

$\displaystyle \ Let \ u \ = \ sinx.$

$\displaystyle \ Then \ du \ = \ cosx \ dx.$

\(\displaystyle Then \ \int{u \ du} \ = \ \int{sinx(cosx \ dx) \ = \ what \ ?\)

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$\displaystyle C) \ \ The \ derivative \ of \ cosx \ = \ -sinx.$

$\displaystyle Let \ u \ = \ cosx.$

$\displaystyle Then \ du \ = \ -sinx \ dx.$

\(\displaystyle Then \ \int{u \ du} = \ \int{cosx(-sinx \ dx)\)

$\displaystyle But \ your \ original \ function \ does \ not \ have \ a \ negative \ sign, \ so$

\(\displaystyle (-1) \int{cosx[(-1)(-sinx \ dx)] \ = \\)

$\displaystyle -\int{cosx(sinx \ dx)} = \ what \ ?$