7.3.18, 7.3.25, and help on 7.3.30

7.3.18
For simplicity, I will assume $a$ and $b$ or positive real numbers.
$$\int \frac{dx}{[(ax)^2-b^2]^{3/2}}$$
We identify that we should use secant.
Let $ax=b\sec\theta$. Then $a dx=b\sec\theta\tan\theta d\theta$
We obtain $$\int \frac{\frac{b}{a}\sec\theta \tan\theta d\theta}{[(b\sec\theta)^2-b^2]^{3/2}}$$
Simplifying the denominator we get $$\int \frac{\frac{b}{a}\sec\theta \tan\theta d\theta}{[b\tan\theta]^{3}}$$
Thus we get $$\int \frac{\sec\theta d\theta}{a\cdot b^2 \tan^2 \theta}$$
If we were to use sines and cosines and cancel wherever possible we'd get
$$\int \frac{\cos\theta d\theta}{a\cdot b^2 \sin^2\theta }$$
Now we do a $u$-substitution. Let $u=\sin\theta$. Then $du=\cos\theta d\theta$.
$$\int \frac{du}{a\cdot b^2 u^2}=\frac{-1}{a\cdot b^2 \cdot u}$$
Substituting back in we get $$\frac{-1}{a \cdot b^2 \cdot \sin\theta}$$
To solve for $\theta$ we can use triangles or we can use trigonometric identities. We change our original transformation to $\cos \theta = \frac{b}{ax}$. Then we know $\sin^2 \theta+\cos^2\theta=1$, so $\sin\theta=\sqrt{1-\cos^2 \theta}=\sqrt{1-(\frac{b}{ax})^2}$.
We conclude our integral is $$\frac{-x}{b^2 \sqrt{(ax)^2 -b^2 )}}$$

Let's check this.
$$\frac{-1}{b^2 \sqrt{(ax)^2 -b^2 )}}+\frac{(ax)^2}{b^2 \sqrt{(ax)^2 -b^2 )}^3}$$
$$\frac{-(ax)^2+b^2}{b^2 \sqrt{(ax)^2 -b^2 )}^3}+\frac{(ax)^2}{b^2 \sqrt{(ax)^2 -b^2 )}^3}$$
$$\frac{b^2}{b^2 \sqrt{(ax)^2 -b^2 )}^3}$$
$$\frac{1}{ \sqrt{(ax)^2 -b^2 )}^3}$$
Yay!

7.3.25 See Example 7
$$\int \frac{x}{\sqrt{x^2+x+1}} dx$$
Complete the square:
$x^2+x+1=(x^2+x+\frac{1}{4})-\frac{1}{4}+1=(x+\frac{1}{2})^2+\frac{3}{4}$
Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$
Then $dx=\frac{\sqrt{3}}{2}\sec^2\theta d\theta$
Then $$\int \frac{\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}}{\frac{\sqrt{3}}{2}\sec\theta} \frac{\sqrt{3}}{2}\sec^2\theta d\theta$$
Cancelling terms we get $$\int \left[ \frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}\right] \sec\theta d\theta$$
The integral of $\tan\theta \sec\theta$ is $\sec\theta$.
By now you should have memorized the formula on page 475 that gives you the integral of $\sec \theta$ to be $\ln \left| \sec \theta + \tan \theta \right| + C$.

Speaking of which. The other day we were brainstorming and we got to $\frac{1}{\sin(x)}$. I now realize this is after all, $\sec{x}$ from which we can use the formula on page 475.

7.3.30
I saw a straightforward way to do the problem, but in the end it seems to be prone to error. So I suggest first doing it the "easier" way. That is, make a $u$-substitution first. Let $u=\sin t$. Then $du=\cos t dt$.
Thus $$\int _0 ^{\pi/2} \frac{\cos t}{\sqrt {1+\sin^2 t }}dt=\int_{t=0}^{t=\pi/2}\frac{du}{\sqrt{1+u^2}}$$
From here we use what we learned in 7.3