4
$\begingroup$

I'm working on an integration by parts problem, and I'm trying to substitute to simplify the equation:

$\int_\sqrt{\frac{\pi}{2}}^\sqrt{\pi} \theta^3 \cos(\theta^2) d\theta$

Using the substitution rule for definite integrals, I substitute $\theta^2 = t$ and apply the same to the limits of integration:

$\int_\frac{\pi}{2}^\pi t^\frac{3}{2} \cos(t) dt$

However, Wolfram|Alpha tells me that I have done something wrong, as these two integrals are not equivalent. Where did I screw up?

  • 3
    When you do a substitution, in addition to changing the limits and all the instances of the variable, you also have to take care of the "differential". Here, $d\theta$; you don't just switch it to a $dt$.2011-02-22

2 Answers 2

3

Here's how I tend to think about substitution in integrals: you've decided to let $t=\theta^2$. The corresponding relationship in differentials is $dt=2\theta d\theta$. Now, back to the integral, try to rearrange the parts of the integral into recognizable chunks. $\begin{align} \int_\sqrt{\frac{\pi}{2}}^\sqrt{\pi} \theta^3 \cos(\theta^2) d\theta &=\int_{\theta=\sqrt{\frac{\pi}{2}}}^{\theta=\sqrt{\pi}} \theta^2 \cos(\theta^2)\frac{1}{2}\cdot 2\theta d\theta \\ &=\frac{1}{2}\int_{t=\frac{\pi}{2}}^{t=\pi} (\theta^2) \cos(\theta^2)(2\theta d\theta) \\ &=\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi} t \cos(t)dt \end{align}$ In particular, notice that the $d\theta$ doesn't just magically become $dt$; rather, $dt$ replaces the equivalent expression $2\theta d\theta$.

0

We have $dt = 2 \theta \ d \theta$. So you have to multiply by $\frac{1}{2}$.

  • 4
    That I agree with, but the integral with respect to $t$ should not have $\theta^3$, it should have $\theta^2=t$, because, as you mentioned, $dt=2\theta d\theta$.2011-02-22