Key takeaway: Sometimes we need to multiply or divide the entire integral by a constant, so we can achieve the appropriate form for u -substitution without changing the value of the integral.

Performing u -substitution with definite integrals is very similar to how it's done with indefinite integrals, but with an added step: accounting for the limits of integration.

𝘶-substitution: indefinite integrals Google Classroom Microsoft Teams ∫ 3 x 2 (x 3 + 1) 6 d x =

This top one, you still didn't mess up by just setting u equal to sine of 5x, we just have to do an extra substitution in order to work it through all the way. And I was able to do this video despite …

In these series of videos (U-substitution) you introduce the treatment of the derivative operators (dx, du, etc) as fractions. You specify that they really are not, but treat them like that anyway.

For example, if it was a natural log of 0.5 or, who knows, whatever it might be. But then we are all done. We have simplified what seemed like a kind of daunting expression.

Some integrals like sin (x)cos (x)dx have an easy u-substitution (u = sin (x) or cos (x)) as the 'u' and the derivative are explicitly given. Some like 1/sqrt (x - 9) require a trigonometric ratio to …

So let's just start with an example here, so let's say we wanna take the indefinite integral of two x plus one times the square root of x squared plus x, dx, does u substitution apply here and if it …

When using 𝘶-substitution in definite integrals, we must make sure we take care of the boundaries of integration.

U substitution is like the reverse chain rule. When we used the derivative chain rule, we saw that d/dx f (u (x)) = df (u)/du*du/dx. U substitution exploits this pattern in evaluating indefinite integrals.