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Help with this small integral step..
- Thread starter OrangeOne
- Start date
D
Deleted member 4993
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OrangeOne said: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:
\(\displaystyle sin(x)\cdot cos(x) \ = \ \frac{1}{2}\cdot sin(2x)\)
OrangeOne said:? 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)] \ = \\)
\(\displaystyle -\int{cosx(sinx \ dx)} = \ what \ ?\)