Need better explanation for part of trig substitution problem with intengral

Question

I've been stuck on this problem for a long time, and have finally managed to get all but the part with the.... bounds? of the integral figured out. I can't seem to get my head around what's going on to take the $\frac{3\sqrt{3}}{2}$ and $0$ to a $1/2$ and $1$. Could someone please try and explain what's going on there a bit better?

Answer 1

Basically, the general case is; if you have an integral $\int_a^bf(x) dx$ and substitute $x$ by $u(x)$, then the bounds are replaced by, respectively, $u(a)$ and $u(b)$.

Let's look at the first substitution we make: $x=\frac{3}{2}\tan \theta$. Then $\theta=\tan^{-1}(\frac{2}{3}x)$. So the bounds $x=0$ and $x=3\sqrt 3/2$ become $\theta=0$ and $\theta=\tan^{-1}\sqrt 3=\pi/3$.

As for the second one, we set $u=\cos \theta$ $ \big(=\cos (\tan^{-1}(\frac{2}{3}x))\big)$. Since we already know the bounds for the $\theta$-integral, we have $\int_0^{3\sqrt 3/2}... dx=\int_0^{\pi/3}... d\theta=\int_{\cos 0}^{\cos(\pi/3)}... du=\int_1^{1/2}... du$

Answer 2

The first integral has the dummy variable $x$ which goes from $0$ to $\frac{3\sqrt 3}2$. The substitution $x=\frac 32\tan\theta$ is then made, so the limits are now for $\theta$ from

$$x_1=0=\frac 32\tan\theta_1$$ $$0=\tan\theta_1$$ $$\theta_1=0$$

to

$$x_2=\frac{3\sqrt 3}2=\frac 32\tan\theta_2$$ $$\sqrt 3=\tan\theta_2$$ $$\theta_2=\frac{\pi}3$$

So the next integral has $\theta$ from $0$ to $\frac{\pi}3$.

The next substitution is $u=\cos\theta$. So the limits are now for $u$ from

$$u_1=\cos\theta_1=\cos 0$$ $$u_1=1$$

to

$$u_2=\cos\theta_2=\cos\frac{\pi}3$$ $$u_2=\frac 12$$

So the next integral has $u$ from $1$ to $\frac 12$.

These all agree with what you see in your text. One strange thing is that the last integral, for $u$, runs from a larger value to a smaller value. This is due to the substitution $u=\cos\theta$, since the cosine function is a decreasing function in the first quadrant: a larger $\theta$ corresponds to a smaller $\cos\theta$. The rules of calculus tell us that is just fine, and everything will work out fine in the end. As it does in your case: the integral still ended up being positive.

Another strange thing is a difference in the two substitutions. The first substitution $x=\frac 32\tan\theta$ gives the old variable in terms of the new, while the second substitution $u=\cos\theta$ gives the new variable in terms of the old. Therefore the methods of finding the new bounds differ, as you see in my equations above. Again, this is legitimate, if confusing, since both functions from old to new variable are one-to-one (at least in the ranges used, in the first quadrant), so we can describe the relationships between the variables in either direction. Doing the substitutions in the way shown in your text made the integrals easier, though understanding the new bounds became harder.

Is it all clear now?