Integral Calculation Using Trig Substitutions

[DeadWolfe]

Our professor today was showing us how to find integrals using trig subsitions.

I was wonder, in a substituation, say, like, x = sin u, how are we entitled to make such a substitution without restricting the values of x, when the sin function's values are between -1 and 1.

 

[learningphysics]

Our professor today was showing us how to find integrals using trig subsitions.

I was wonder, in a substituation, say, like, x = sin u, how are we entitled to make such a substitution without restricting the values of x, when the sin function's values are between -1 and 1.

Yes, the values of x are restricted. Hopefully the limits of integration will be between x=-1 and x=1

You will know that the substitution is invalid when you try to determine the new limits of integration. If you have an integral that goes from x=3 to x=5,
and you substituted x=sin u, you'll know that you've made a mistake since 3=sinu and 5=sinu have no solutions.

 

[M98Ranger]

Great question! When using trig substitutions, we are essentially changing the variables in our integral to make it easier to solve. In the example you mentioned, x = sin u, we are essentially replacing x with sin u in our integral. This substitution is valid because the values of x and sin u are related through the trigonometric identity x = sin u. This means that for any value of x, there exists a corresponding value of sin u and vice versa. So even though the values of sin u are restricted between -1 and 1, we can still use this substitution in our integral because it is a valid mathematical relationship. Hope that helps clarify things!