A review of integration techniques

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We review common techniques to compute indefinite and definite integrals.

In some sense, calculating derivatives is straightforward. Most of the functions of interest are sums, scalar multiples, products, quotients, or compositions of common functions whose derivatives we can find. A “brute force” application of the sum, scalar multiple, product, quotient, and chain rules will eventually lead to a correct derivative. Integration is more of an art form. We often have to apply various techniques in order to write down a nice formula for the antiderivatives. We explore some of the common ones here.

1 Preliminary algebraic simplification

A good first step in attempting to compute antiderivatives involves simplifying the integrand first. Doing so requires careful algebra.

Compute:

\[ \int (7-x^2)^2 dx \]

Note first that $(7-x^2)^2 \neq 49-x^4$; we need to expand the integrand by writing out $(7-x^2)(7-x^2)$:

\[ \int (7-x^2)^2 dx = \int \answer [given]{49 - 14x^2 + x^4} dx \]

It’s now not difficult to finish the problem:

\[ \int (49 - 14x^2 + x^4) dx = \answer [given]{49x - \frac {14 x^3}{3} + \frac {x^5}{5}} + C \]

Consider the indefinite integral $\int \frac {2x^2-3}{2x^2} dx$.

While it may be tempting, note that $\frac {2x^2-3}{2x^2} \neq \frac {\cancel {2x^2}-3}{\cancel {2x^2}}$. We can never simplify by cancelling terms over addition or subtraction in the denominator. Instead, we can split the fraction up then simplify. \begin{align*} \int \frac {2x^2-3}{2x^2} dx &=\int \frac {2x^2}{2x^2}-\frac {3}{2x^2} dx \\ &= \int 1-\frac {3}{2}x^{-2} dx \\ &= x+ \answer [given]{\frac {3}{2}x^{-1}}+C \end{align*}

2 Substitution

Sometimes, it is helpful to try to recognize when an integral involves reversing the chain rule for differentiation. When an integrand involves a composition of a trigonometric, exponential, or power and another function, letting a new variable represent the inner function helps.

Consider $\int 2x e^{x^2+1} dx$.

Note that no algebra will help us simplify the expression. The only antidifferentiation formula we have regarding the exponential is $\int e^x dx = e^x +C$. Let’s start by letting the exponent in the original integral be a new variable $u$:

Let $u= x^2+1$. We now need to write everything in the integral in terms of this new variable. Recalling that if $u = f(x)$, the differentials are related by $du = f'(x) dx$, we can write $du = \answer [given]{2x} dx$.

We thus find

The antiderivative in $u$ is now easy to compute.

\[ \int e^u du = \answer [given]{e^u}+C \]

Reversing the substitution by setting $u=x^2+1$ above now gives $\int 2xe^{x^2+1} dx = e^{x^2+1}+C$.

Note that when we change the variable of integration, it is necessary to replace both the original variable and the differential. That is, in the above example, we need to replace all $x$ terms with $u$ and write $dx$ in terms of $u$ and $du$.

In the last example, after letting $u$ be the exponent and transforming $dx$, there were no $x$ terms left. This does not always happen, but it does not mean that substitution does not aid in computing the indefinite integral.

Compute $\int 4x \sqrt {2x+1} dx$.

Here, there is no preliminary algebra that will be helpful. Since we have a square root of a function of $x$, we set $u=2x+1$ and find $du = \answer [given]{2} dx$. Thus:

\[ \int 4x \sqrt {2x+1} dx = \int 4x \sqrt {u} \cdot \frac {du}{2} = \int 2x \sqrt {u} du \]

Note that we now have an extra $x$ term in the integrand. Since $x$ depends on $u$, we cannot move it outside of the integral. Before abandoning hope that our substitution is useful here, we can ask whether it is easy to express this leftover $x$ in terms of $u$.

Since $u= 2x+1$, we find that $x = \answer [given]{\frac {1}{2}(u-1)}$, so we can now rewrite the integral as:

\[ \int 4x \sqrt {2x+1} dx = \int 2x \sqrt {u} du = \int 2 \cdot \frac {1}{2}(u-1) \sqrt {u} du = \int (u-1)\sqrt {u} du \]

We can now perform some algebra, then integrate.

\[ \int (u-1)\sqrt {u} du = \int \left (u^{3/2} - u^{1/2} \right ) du = \answer [given]{\frac {2}{5} u^{5/2} - \frac {2}{3} u^{3/2}} +C \]

Substituting $u=2x+1$ to obtain the antiderivatives in terms of $x$, we find that

\[ \int 4x \sqrt {2x+1} dx = \frac {2}{5} (2x+1)^{5/2} - \frac {2}{3} (2x+1)^{3/2} +C. \]

Let’s see another example involving a trickier substitution and a definite integral. First, recall the substitution theorem for definite integrals:

Integral Substitution Formula If $u$ is differentiable on the interval $[a,b]$ and $f$ is differentiable on the interval $[u(a),u(b)]$, then

\[ \int _a^b f'(u(x)) u'(x) dx =\int _{u(a)}^{u(b)} f'(u) du. \]

This may look difficult to read, but it really just reminds us that to work with definite integrals, we need to write the integrand, the differential, and the limits of integration in terms of the new variable.

Compute $\int _0^2 \frac {6x^2}{(x^3+1)^2} dx$

We begin by setting $u = \answer [given]{x^3+1}$. We now have to write the limits of integration in terms of $u$ as well as express the differential $dx$ in terms of $u$ and $du$.

For the limits of integration, note that when $x=0$, we have $u = (0)^3+1 = 1$. Similarly, when $x=2$, $u=\answer [given]{9}$.

Since $u=x^3+1$, we find $du = \answer [given]{3x^2} dx$.

We can now write our original integral in terms of $u$:

\[ \int _{x=0}^{x=2} \frac {6x^2}{(x^3+1)^2} dx = \int _{u=\answer [given]{1}}^{u=\answer [given]{9}} \answer [given]{\frac {2}{u^2}} du \]

We can now evaluate the integral:

\[ \int _{u=1}^{u=9} \frac {2}{u^2} du = \int _{u=1}^{u=9}2u^{-2} du = \bigg [-2u^{-1}\bigg ]_{u=1}^{u=9} = -\frac {2}{9} - (-2) = \frac {16}{9} \]

3 Splitting Up Fractions

In a similar vein, if you have a rational function, it can help to separate the integrand into several fractions.

Compute:

\[ \int \frac {6x-8}{1+x^2} dx \]

Note that the substitution $u= 1+x^2$ is tempting, but it only will help us with part of the integrand! Write with me:

\[ \int \frac {6x-8}{1+x^2} dx = \int \frac {6x}{1+x^2}-\frac {8}{1+x^2} dx \]

Using a substitution, $ \int \frac {6x}{1+x^2} dx = \answer [given]{3 \ln (1+x^2)}$+C, and using the inverse tangent antidifferentiation formula, $ \int \frac {8}{1+x^2} dx = \answer [given]{8 \arctan (x)}$+C.

Hence, $\int \frac {6x-8}{1+x^2} dx = \answer [given]{3 \ln (1+x^2)- 8 \arctan (x)}+C$

Compute:

\[ \int \frac {\cos ^2(x)+1}{3 \cos ^2(x)} dx \]

It may be tempting to try to cancel just the expressions involving $\cos ^2(x)$, but we have to split up the entire numerator over the denominator.

\begin{align*} \int \frac {\cos ^2(x)+1}{3 \cos ^2(x)} dx &= \int \frac {\cos ^2(x)}{3 \cos ^2(x)} + \frac {1}{3 \cos ^2(x)} dx \\ &= \int \frac {1}{3} + \frac {1}{3} \sec ^2(x) dx \\ & = \answer [given]{\frac {1}{3}x+\frac {1}{3}\tan (x)}+C \end{align*}

4 Final Thoughts

Understanding how to integrate is highly dependent on having a good grasp on how to differentiate functions and being comfortable with algebra. The techniques presented here are not an exhaustive list. In order to become proficient computing integrals, there is no substitute for practice.

2025-01-06 18:16:00