Evaluate $\displaystyle \int \frac{1}{x^2 \sqrt{x^2 - 9}} \ dx $
That part under the square root should tell us what trigonometric substitution to use here. What is that?
Use the substitution $x = 3 \sec{\theta}$ because $\sec^2(\theta) - 1 = \tan^2(\theta)$
The square root in the denominator becomes $3 \tan \theta$, and since $dx = 3 \sec{\theta} \tan {\theta}$ some things will cancel.
Once you're done, use a triangle to switch this back to being in terms of $x$.
$\displaystyle \frac{\sqrt{x^2 - 9}}{9x} + C$
Evaluate $\displaystyle \int \frac{1}{x^2 \sqrt{25 - x^2}} \ dx $
That part under the square root should tell us what trigonometric substitution to use here. What is that?
Use the substitution $x = 5 \sin{\theta}$ to make this work out.
Make sure to also change $dx$ to $d\theta$ and use that to cancel things out.
Once you're done, use a triangle to switch this back to being in terms of $x$.
$\displaystyle - \frac{\sqrt{25-x^2}}{25x} + C $
Compute $\displaystyle \int x^3 \sqrt{36 - x^2} dx$
That part under the square root should tell us what trigonometric substitution to use here. What is that?
Use the substitution $x = 6 \sin{\theta}$ to make this work out.
Once you simplify this out, it falls into the trigonometric integrals category. Thankfully, this one is an easy one.
You need to evaluate $\displaystyle 6^5 \int \sin^3(\theta) \cos^2(\theta)\ d\theta$ and then convert back to $x$.
$\displaystyle \frac{1}{5} (36-x^2)^{5/2} - 12 (36-x^2)^{3/2} + C$
Compute $\displaystyle \int \frac{1}{y^3 \sqrt{y^2 - 16}}\ dy$
That part under the square root should tell us what trigonometric substitution to use here. What is that?
Use the substitution $y = 4 \sec{\theta}$ to solve this integral, because the square root on the bottom becomes $4 \tan{\theta}$.
The expression that needs to be simplified is $\displaystyle \frac{4\sec{\theta} \tan{\theta}}{4^3 \sec^3{\theta} 4 \tan{\theta}}$.
The resulting integral will be $\frac{1}{64} \int \cos^2{\theta}\ d\theta $, and when you have $\sin(2\theta)$, you'll want to rewrite it as $2 \sin(\theta) \cos(\theta)$ so that you can convert it back into $x$.
$\displaystyle \frac{1}{128} \sec^{-1}(\frac{y}{4}) + \frac{\sqrt{y^2 - 16}}{16y^2} + C$
Compute $\displaystyle \int \frac{1}{(121-x^2)^2}\ dx $
This may not look like trigonometric substitution at first, but it turns out that this is a good way to solve this integral.
Try the substitution $x = 11 \sin(\theta)$ to solve out this integral.
The resulting integral you want to solve should be $\displaystyle \frac{1}{1331} \int \sec^3{\theta}\ d\theta$
This will need the reduction formulas and the fact that the antiderivative of $\sec(\theta)$ is $\ln |\sec{\theta} + \tan{\theta}|$ to finish up the problem.
$\displaystyle \frac{1}{2662}\left( \frac{11x}{121-x^2} + \ln \left| \frac{11+x}{\sqrt{121-x^2}} \right| \right) + C$
Compute $\displaystyle \int \sqrt{x^2 + 6x} \ dx $
This also does not look like trigonometric substitution right off the bat, but there's no other way to solve this problem. You'll need to complete the square first.
The part under the square root can be rewritten as $(x+3)^2 - 9$, so that the appropriate trigonometric substitution is $x+3 = 3\sec{\theta}$.
This substitution will result in needing to solve a trigonometric integral, which will need reduction formulas.
You end up needing to solve $\displaystyle 9 \int \sec^3(\theta) - \sec(\theta)\ d\theta$ and then convert back to $x$.
$\displaystyle \frac{(x+3)(\sqrt{x^2 + 6x})}{2} - \frac{9}{2} \ln \left(\frac{x+3+\sqrt{x^2 + 6x}}{3} \right| + C$
Compute $\displaystyle \int \frac{1}{(4 - x^2)^{3/2}}\ dx$.
Compute $\displaystyle \int_0^1 \frac{dt}{(t^2 + 9)^2} $