Thread: Integration Limits

My question is as follows: In learning integration, I was taught to change the limits of integration when using u-substitution and solve accordingly. However, now that I am solving definite integrals involving inverse trigonometric functions, my calculus book is no longer making substitutions for the limits of integration based on u-substitutions that I am making? Please explain..........

Originally Posted by kaiser0792

My question is as follows: In learning integration, I was taught to change the limits of integration when using u-substitution and solve accordingly. However, now that I am solving definite integrals involving inverse trigonometric functions, my calculus book is no longer making substitutions for the limits of integration based on u-substitutions that I am making? Please explain..........

Try posting the actual questions.

It is not necessary to change the limits, you can always convert the function back to a function of $\displaystyle x$ and then substitute the $\displaystyle x$ values.

But my problem is that when the book was teaching u-substitution, I understood that unless limits of integration were changed that the final answer would have to be changed back in terms of x. However, in the chapter on integration of inverse trig functions, the book uses u-substitution but doesn't change the limits of integration in terms of u and just integrates as if the integral was written in terms of x. So how is it that in some instances, u-substitution warrants changes in limits of integration, and in other instances, u-substitution is used but the limits of integration are not changed to compensate for the u-substitution?

Originally Posted by kaiser0792

But my problem is that when the book was teaching u-substitution, I understood that unless limits of integration were changed that the final answer would have to be changed back in terms of x. However, in the chapter on integration of inverse trig functions, the book uses u-substitution but doesn't change the limits of integration in terms of u and just integrates as if the integral was written in terms of x. So how is it that in some instances, u-substitution warrants changes in limits of integration, and in other instances, u-substitution is used but the limits of integration are not changed to compensate for the u-substitution?

So like I said, post the specific problem with what the book has instructed, and we will be able to explain why they didn't change the limits with the substitution.

Definite Integral evaluated from 0 to 1/6: 1/(1-9x^2)^1/2 dx requires u-substitution of u=3x, but the final answer does not reflect a change in limits of integration from 0 to 1/2 based on the u-substitution of u=3x. The integral is evaluated from 0 to 1/6 even though u-substitution was employed and the answer is pi/18. So again, my question is this: in other instances, the limits of integration would be changed to (0 to 3(1/6) or (0 to 1/2) based on the u-substitution, but in this case they are not changed and the integration is done based on the original limits of integration even thought u-substitution was used?????

Originally Posted by kaiser0792

Definite Integral evaluated from 0 to 1/6: 1/(1-9x^2)^1/2 dx requires u-substitution of u=3x, but the final answer does not reflect a change in limits of integration from 0 to 1/2 based on the u-substitution of u=3x. The integral is evaluated from 0 to 1/6 even though u-substitution was employed and the answer is pi/18. So again, my question is this: in other instances, the limits of integration would be changed to (0 to 3(1/6) or (0 to 1/2) based on the u-substitution, but in this case they are not changed and the integration is done based on the original limits of integration even thought u-substitution was used?????

Did they convert the integral back to a function of $\displaystyle x$ before they substituted the limits?

If you are saying they integrated $\displaystyle \int_0^{1/6} \frac{dx}{\sqrt{1- 9x^2}$ by making the substitution u= 3x so that du= 3 dx and then writing it as
$\displaystyle \frac{1}{3}\int_0^{1/6}\frac{du}{\sqrt{1- u^2}}= \frac{1}{3}\left[ sin^{-1}(u)\right]_0^{1/6}= \frac{1}{3}sin^{-1}(1/6)$
which is about 0.05582, then that is just wrong.

Correct is either
$\displaystyle \frac{1}{3}\int_0^{1/2} \frac{du}{\sqrt{1- u^2}}= \frac{1}{3}\left[ sin^{-1}(u)\right]_0^{1/2}= \frac{1}{3} sin^{-1)(1/2)$ which is about 0.17453

or replacing [tex]sin^{-1}(u)[tex] with $\displaystyle sin^{-1}(3x)$ to get $\displaystyle \left[\frac{1}{3}sin^{-1}(3x)\right]_0^{1/6}= sin^{-1}(3(1/6)= sin^{-1}(1/2)$ and again, about 0.17453.