The Notes: `int t sin^2(t) dt` Evaluate the integral

Thursday, April 1, 2010

$\displaystyle\int{t}{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$ Evaluate the integral

$\displaystyle\int{t}{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$

Let's use the method of integration by parts,

If f(x) and g(x) are differentiable functions, then

$\displaystyle\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}{\left.{d}{x}\right.}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

If we write f(x)=u and g'(x)=v, then

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Let's denote u=t and $\displaystyle{v}={{\sin}}^{{2}}{\left({t}\right)}$

$\displaystyle{u}'=\frac{{d}}{{\left.{d}{t}}}{\left({t}\right)}={1}$

$\displaystyle\int{t}{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}={t}\int{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}-\int{\left(\int{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}\right)}{\left.{d}{t}\right.}$

Noe let's use the identity : $\displaystyle{{\sin}}^{{2}}{\left({x}\right)}=\frac{{{1}-{\cos{{\left({2}{x}\right)}}}}}{{2}}$

$\displaystyle={t}\int\frac{{{1}-{\cos{{\left({2}{t}\right)}}}}}{{2}}{\left.{d}{t}\right.}-\int{\left(\int{\left(\frac{{{1}-{\cos{{\left({2}{t}\right)}}}}}{{2}}\right)}{\left.{d}{t}\right.}\right)}{\left.{d}{t}\right.}$

$\displaystyle=\frac{{t}}{{2}}\int{\left({1}-{\cos{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}-\int{\left(\frac{{1}}{{2}}\int{\left({1}-{\cos{{\left({2}{t}\right)}}}\right)}{\left.{d}{t}\right.}\right)}{\left.{d}{t}\right.}\right.}$

$\displaystyle=\frac{{t}}{{2}}{\left({t}-\frac{{\sin{{\left({2}{t}\right)}}}}{{2}}\right)}-\frac{{1}}{{2}}\int{\left({t}-\frac{{\sin{{\left({2}{t}\right)}}}}{{2}}\right)}{\left.{d}{t}\right.}$

$\displaystyle=\frac{{t}}{{2}}{\left({t}-\frac{{\sin{{\left({2}{t}\right)}}}}{{2}}\right)}-\frac{{1}}{{2}}{\left(\frac{{{t}}^{{2}}}{{2}}-{\left(-\frac{{\cos{{\left({2}{t}\right)}}}}{{4}}\right)}\right)}$

$\displaystyle=\frac{{{t}}^{{2}}}{{2}}-\frac{{{t}{\sin{{\left({2}{t}\right)}}}}}{{4}}-\frac{{{t}}^{{2}}}{{4}}-\frac{{\cos{{\left({2}{t}\right)}}}}{{8}}$