What are some Integration Techniques unrelated to the Antiderivative? Are there any definite integration techniques which I could learn (calc AB student)? I mean techniques which don't require you to find the anti derivative. Thanks!
 A: All of the following apply mostly to definite integrals and not so much to indefinite integrals:
1) First off, geometry!  Some integrals are best done by noting they represent some well known geometric shape, such as the circle, or a cone.  Rather basic of course, but useful sometimes. e.g.
$$\int_a^b\frac1{\sqrt{1-x^2}}\ dx$$
This is simply the arc length of a circle, though you might not have noticed immediately, and it can be solved by taking a part of the circumference and dividing appropriately.
2) Symmetry: As you have noticed, some integrals can be tackled with symmetry, especial integrals involving trig or some nasty polynomials and things of that nature.  e.g.
$$\int_0^{\pi/2}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}\ dx$$
Notice that taking the integral backwards from $\pi/2$ to $0$, you get
$$\int_0^{\pi/2}\frac{\cos^n(x)}{\sin^n(x)+\cos^n(x)}\ dx$$
And by adding them together,
$$2I=\int_0^{\pi/2}\frac{\sin^n(x)+\cos^n(x)}{\sin^n(x)+\cos^n(x)}\ dx=\frac\pi2$$
This is best done by graphing the integrand and seeing if you can note anything special, or if the function looks like a puzzle piece to another integral, then proving what you think to be the case rigorously.
3) Differentiation under the integral sign (DUIS): It is what it sounds like.  You take some integral, and then take the derivative of it, though probably not in the way you'd expect.  e.g.
$$f(x)=\int_0^1\frac{t^x-1}{\ln(t)}\ dt$$
This converges for any $x\ge0$, thus, we may take the derivative with respect to $x$:
$$f'(x)=\int_0^1t^x\ dt=\frac1{1+x}$$
Integrating back:
$$f(x)=c+\ln(1+x)$$
And by setting $x=0$, we find $c=0$, thus
$$\int_0^1\frac{t^x-1}{\ln(t)}\ dt=\ln(1+x)$$
5) Polar coordinates.  e.g.
$$I=\int_{-\infty}^{+\infty}e^{-x^2}\ dx$$
Notice then that
$$I^2=\int_{-\infty}^{+\infty}e^{-x^2}\ dx\int_{-\infty}^{+\infty}e^{-y^2}\ dy=\iint_{\mathbb R^2}e^{-(x^2+y^2)}\ dx\ dy$$
This represents an integral over $\mathbb R^2$, but after converting to polar and letting $u=r^2$, this becomes
$$I^2=\int_0^{+\infty}2\pi re^{-r^2}\ d\theta\ dr=\int_0^{+\infty}\pi e^{-u}\ du=\pi$$
Noting that the $2\pi r$ comes from circumference.  We finally conclude that
$$\int_{-\infty}^{+\infty}e^{-x^2}\ dx=\sqrt\pi$$
6) There exist other more advanced integral transforms, and I won't go into all of them, but the basic idea is simple: You apply a transformation on your integral to get a new integral, and then apply the inverse transform to recover your integral.  Convolutions are similarly useful here.
7) Complex analysis: Basically, this extends the idea of integration to the complex plane, and rather than integrating "along the real line from point a to point b" we may now integrate "along some path $\gamma$ in the complex plane from point a to point b".  Immediate advantages include being able to exploit more symmetries, taking advantage of real and imaginary parts, taking advantage of branch cuts, applying the residue theorem, and Cauchy's integral formula.  e.g.
$$I=\int_0^{2\pi}e^{\cos(x)}\cos(x-\sin(x))\ dx$$
Notice this to be the real part of another integral:
$$I=\Re\int_0^{2\pi}e^{\cos(x)+i(x-\sin(x))}\ dx=\Re\int_0^{2\pi}e^{ix}e^{1/e^{ix}}\ dx$$
By letting $z=e^{ix}$, and noting this becomes an integral along $|z|=1$ counterclockwise,
$$I=\Re\oint_{|z|=1}\frac1ie^{1/z}\ dz$$
And finally by the residue formula,
$$I=2\pi$$
8) Series expansions: Even though it doesn't always omit a solution, at least it provides a form good for numerical approximations.  Here is a good example:
$$I=\int_0^{+\infty}\frac x{e^x-1}\ dx$$
We then note that $\frac x{e^x-1}=\frac{xe^{-x}}{1-e^{-x}}$ and apply the geometric series to get
$$I=\int_0^{+\infty}xe^{-x}+xe^{-2x}+xe^{-3x}+\dots\ dx$$
And by integrating term by term,
$$I=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\dots$$
which is the well known Basel problem (or $\zeta(2)$ if you are more familiar with that) and gives us
$$\int_0^{+\infty}\frac x{e^x-1}\ dx=\frac{\pi^2}6$$
