[integral]sin^2(x)*dx -> how do get its antiderivative ?

[lucky_luke]

hi

How did we get antiderivative

 1/2*x - cos(x)*sin(x)

from

[integral]sin^2(x) * dx

?


thank you

 

[galactus]

By noting that \(\displaystyle sin^{2}(x)=\frac{1}{2}(1-cos(2x))\)

\(\displaystyle \L\\\frac{1}{2}\int{1-cos(2x)}dx\)

 

[lucky_luke]

By noting that \(\displaystyle sin^{2}(x)=\frac{1}{2}(1-cos(2x))\)

\(\displaystyle \L\\\frac{1}{2}\int{1-cos(2x)}dx\)


I don't believe the antiderivative you have posted is the antiderivative of

\(\displaystyle sin^{2}(x)\)

I forgot to add parenthesis, so real antiderivative is 1/2(x - cos(x)*sin(x)).

You can check it at this url:

http://integrals.wolfram.com/index.jsp

So you have any idea how they got that result?

 

[pka]

\(\displaystyle \L
\begin{eqnarray}
\int {\sin ^2 (x)dx = \int {\left( {\frac{1}{2}} \right)\left[ {1 - \cos (2x)} \right]dx} } \\
= \left( {\frac{1}{2}} \right)\left[ {x - \left( {\frac{1}{2}} \right)\sin (2x)} \right] \\
= \left( {\frac{1}{2}} \right)\left[ {x - \sin (x)\cos (x)} \right] \\
\end{array}\)

 

[galactus]

You're so right, pka. Oversight on my part...DUH!!. Thanks for the correction.