Difference between revisions of "Calc1.51"

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[[Main Page]] : [[Calculus]]
 
[[Main Page]] : [[Calculus]]
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Test
 
<math>\int \sin u du\,</math>
 
<math>\int \sin u\,du\,</math>
 

Latest revision as of 13:33, 27 February 2006

\int x^{2}\sin x^{3}\,dx\,

We know how to do the integral of \sin x\, so a good substitution to try would be u=x^{3}\, because if it works out we will be left with \sin u\, which we can integrate.

u=x^{3}\,

du=3x^{2}dx\,

dx={\frac  {du}{3x^{2}}}\,

Substitute these into the integral to see if it changes the integral into a form we can integrate.

\int x^{2}\sin x^{3}\,dx=\int x^{2}\sin u{\frac  {du}{3x^{2}}}={\frac  {1}{3}}\int \sin u\,du\,

This is an integral we can compute.

{\frac  {1}{3}}\int \sin u\,du=-{\frac  {1}{3}}\cos(u)+C\,

Finally, undo the substitution

-{\frac  {1}{3}}\cos(u)+C=-{\frac  {1}{3}}\cos(x^{3})+C\,


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