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
    $\begingroup$ 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 $dt$. $\endgroup$ Feb 22, 2011 at 3:07

2 Answers 2


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 $2\theta d\theta$.


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

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    $\begingroup$ Not multiplying by one half is one of two mistakes. The other is the power of t in front of the cosine. $\endgroup$ Feb 22, 2011 at 2:38
  • $\begingroup$ @Jonas Meyer: We have $t^{3/2} = (\theta^{2})^{3/2} = \theta^{6/2}= \theta^3$. $\endgroup$ Feb 22, 2011 at 2:43
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    $\begingroup$ 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, $dt=2\theta d\theta$. $\endgroup$ Feb 22, 2011 at 2:46

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