Integration is a fundamental concept in calculus, and understanding various techniques is essential for solving a wide range of problems, particularly on the AP Calculus AB/BC exam. Here is a comprehensive guide to the primary integration methods you need to master:

1. Basic Antiderivatives

Before diving into more advanced techniques, it’s important to know the basic antiderivatives, which are just the reverse of differentiation. The most common antiderivatives you should remember are:

• ∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C∫xndx=n+1xn+1​+C, for n≠−1n \neq -1n=−1

• ∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C

• ∫sin⁡x dx=−cos⁡x+C\int \sin x \, dx = -\cos x + C∫sinxdx=−cosx+C

• ∫cos⁡x dx=sin⁡x+C\int \cos x \, dx = \sin x + C∫cosxdx=sinx+C

• ∫sec⁡2x dx=tan⁡x+C\int \sec^2 x \, dx = \tan x + C∫sec2xdx=tanx+C

• ∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x \, dx = -\cot x + C∫csc2xdx=−cotx+C

• ∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x \tan x \, dx = \sec x + C∫secxtanxdx=secx+C

• ∫csc⁡xcot⁡x dx=−csc⁡x+C\int \csc x \cot x \, dx = -\csc x + C∫cscxcotxdx=−cscx+C

2. Substitution (u-Substitution)

This is one of the most commonly used techniques, especially when you have a composite function where part of the integrand is a derivative of another part. The general idea is to set a part of the integrand equal to uuu, then replace all instances of the variable in the integral. The steps are:

• Let u=g(x)u = g(x)u=g(x), where g(x)g(x)g(x) is a function inside the integral.

• Find du=g′(x)dxdu = g'(x)dxdu=g′(x)dx, then replace dxdxdx with du/g′(x)du/g'(x)du/g′(x).

• Substitute the new variables into the integral.

• Finally, integrate with respect to uuu, and then substitute back in terms of xxx if necessary.

Example:
∫2xx2+1 dx\int 2x \sqrt{x^2 + 1} \, dx∫2xx2+1​dx

Let u=x2+1u = x^2 + 1u=x2+1, so du=2x dxdu = 2x \, dxdu=2xdx.
The integral becomes:
∫u du=23u3/2+C=23(x2+1)3/2+C\int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C = \frac{2}{3}(x^2 + 1)^{3/2} + C∫u​du=32​u3/2+C=32​(x2+1)3/2+C.

3. Integration by Parts

This method is based on the product rule for differentiation. It’s particularly useful when the integrand is a product of two functions. The formula is:

∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du∫udv=uv−∫vdu

Where:

• uuu is a function you will differentiate.

• dvdvdv is a function you will integrate.

To apply this method:

1. Choose uuu to be the function that simplifies when differentiated (often a logarithmic or polynomial function).

2. Choose dvdvdv to be the function that’s easier to integrate.

3. Differentiate uuu to get dududu, and integrate dvdvdv to get vvv.

4. Apply the integration by parts formula, and if necessary, repeat the process.

Example:

∫xln⁡x dx\int x \ln x \, dx∫xlnxdx

Let u=ln⁡xu = \ln xu=lnx, and dv=x dxdv = x \, dxdv=xdx. Then du=1x dxdu = \frac{1}{x} \, dxdu=x1​dx and v=x22v = \frac{x^2}{2}v=2x2​.

By integration by parts:

∫xln⁡x dx=x22ln⁡x−∫x22⋅1x dx=x22ln⁡x−x24+C\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C∫xlnxdx=2x2​lnx−∫2x2​⋅x1​dx=2x2​lnx−4x2​+C

4. Trigonometric Substitution

This method is useful when you encounter integrals involving square roots of quadratic expressions. By substituting trigonometric functions for parts of the integrand, you can often simplify the integral. Here are the basic forms of substitution:

• For a2−x2\sqrt{a^2 - x^2}a2−x2​, use x=asin⁡θx = a \sin \thetax=asinθ, so dx=acos⁡θ dθdx = a \cos \theta \, d\thetadx=acosθdθ.

• For a2+x2\sqrt{a^2 + x^2}a2+x2​, use x=atan⁡θx = a \tan \thetax=atanθ, so dx=asec⁡2θ dθdx = a \sec^2 \theta \, d\thetadx=asec2θdθ.

• For x2−a2\sqrt{x^2 - a^2}x2−a2​, use x=asec⁡θx = a \sec \thetax=asecθ, so dx=asec⁡θtan⁡θ dθdx = a \sec \theta \tan \theta \, d\thetadx=asecθtanθdθ.

After performing the substitution, the integral becomes a standard trigonometric integral, which you can then solve using basic trigonometric identities.

Example:

∫dx4−x2\int \frac{dx}{\sqrt{4 - x^2}}∫4−x2​dx​

Substitute x=2sin⁡θx = 2 \sin \thetax=2sinθ, so dx=2cos⁡θ dθdx = 2 \cos \theta \, d\thetadx=2cosθdθ. The integral becomes:

∫2cos⁡θ dθ4−4sin⁡2θ=∫ dθ=θ+C\int \frac{2 \cos \theta \, d\theta}{\sqrt{4 - 4 \sin^2 \theta}} = \int \, d\theta = \theta + C∫4−4sin2θ​2cosθdθ​=∫dθ=θ+C

Now, substitute θ=arcsin⁡(x/2)\theta = \arcsin(x/2)θ=arcsin(x/2) to return to xxx.

5. Partial Fraction Decomposition

This technique is used when you have a rational function (a fraction of polynomials) and you want to break it down into simpler fractions. The basic idea is to express the rational function as a sum of simpler fractions whose denominators are factors of the original denominator.

1. Factor the denominator into linear or irreducible quadratic factors.

2. Set up the partial fraction decomposition.

3. Solve for the constants by equating coefficients or substituting convenient values of xxx.

4. Integrate each simple fraction separately.

Example:

∫1(x−1)(x+2) dx\int \frac{1}{(x-1)(x+2)} \, dx∫(x−1)(x+2)1​dx

Decompose as:

1(x−1)(x+2)=Ax−1+Bx+2\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}(x−1)(x+2)1​=x−1A​+x+2B​

Multiply both sides by (x−1)(x+2)(x-1)(x+2)(x−1)(x+2), solve for AAA and BBB, then integrate.

6. Improper Integrals

An improper integral occurs when the interval of integration is infinite or when the integrand approaches infinity at some point in the interval. These integrals can be evaluated by taking limits.

• Infinite Interval: For an integral like ∫a∞f(x) dx\int_{a}^{\infty} f(x) \, dx∫a∞​f(x)dx, you would rewrite it as lim⁡b→∞∫abf(x) dx\lim_{b \to \infty} \int_{a}^{b} f(x) \, dxlimb→∞​∫ab​f(x)dx.

• Unbounded Integrand: For integrals where the function has an infinite discontinuity (like ∫ab1x dx\int_{a}^{b} \frac{1}{x} \, dx∫ab​x1​dx at x=0x = 0x=0), take the limit as the problematic point approaches the bound.

Example:

∫1∞1x2 dx\int_1^\infty \frac{1}{x^2} \, dx∫1∞​x21​dx

Rewrite as lim⁡b→∞∫1b1x2 dx\lim_{b \to \infty} \int_1^b \frac{1}{x^2} \, dxlimb→∞​∫1b​x21​dx, which evaluates to 111.

7. Numerical Integration (Trapezoidal and Simpson’s Rule)

If an integral cannot be solved analytically, numerical methods can be used to approximate the value. The two most common methods are:

• Trapezoidal Rule: Approximates the area under a curve by dividing the region into trapezoids.

∫abf(x) dx≈h2(f(a)+2∑i=1n−1f(xi)+f(b))\int_a^b f(x) \, dx \approx \frac{h}{2} \left( f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right)∫ab​f(x)dx≈2h​(f(a)+2i=1∑n−1​f(xi​)+f(b))

where h=b−anh = \frac{b-a}{n}h=nb−a​ and xix_ixi​ are the intermediate points.

• Simpson’s Rule: Uses parabolas to approximate sections of the curve. It’s more accurate than the trapezoidal rule, especially for smooth functions.

∫abf(x) dx≈h3(f(a)+4∑i=1,3,...,n−1f(xi)+2∑i=2,4,...,n−2f(xi)+f(b))\int_a^b f(x) \, dx \approx \frac{h}{3} \left( f(a) + 4 \sum_{i=1,3,...,n-1} f(x_i) + 2 \sum_{i=2,4,...,n-2} f(x_i) + f(b) \right)∫ab​f(x)dx≈3h​(f(a)+4i=1,3,...,n−1∑​f(xi​)+2i=2,4,...,n−2∑​f(xi​)+f(b))

where h=b−anh = \frac{b-a}{n}h=nb−a​.

Conclusion

Mastering these integration techniques is crucial for success on the AP Calculus AB/BC exam. Knowing when and how to apply each method, along with practice, will make you proficient at solving a wide range of integral problems. Keep practicing, and remember to focus on understanding the underlying principles behind each method for long-term retention and success.