I was messing around proving the area of a circle using trigonometric substitution. However, I ended up with area = -πr^2.(adsbygoogle = window.adsbygoogle || []).push({});

In my integral I ended up using trigonometric substitution and setting x = r*cos(theta)

However, this yields x = -r*sin(theta)*d(theta).

When that's substituted back into my integral, I ended up with the negative value for area. Why is it that if I were to use a different angle theta and express x = r*sin(theta) that I end up with the correct area?

Thanks a lot, as always.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Trigonometric Substitution

Loading...

Similar Threads for Trigonometric Substitution |
---|

B Methods of integration: direct and indirect substitution |

I Derivative When Substituting Variables |

B Not following an integral solution |

**Physics Forums | Science Articles, Homework Help, Discussion**