So, you’re at the exam, you’ve been staring at the question for 10 minutes, and the clock is ticking...How do you know which integration technique to use for a particular question?
If you're enrolled in a calculus II class, you've probably encountered the following techniques:
Basic Substitution Method Integration by Parts Powers of Trigonometric Functions Trigonometric Substitution Partial FractionsIn this blog, we'll explore each of these techniques in greater detail, and consider some examples where the technique applies.
To start, nearly every integrand you come across can be simplified to look like one of these forms:
Most teachers will be kind enough to provide these formulas on a test, either in the form of a derivative or an integral. If not, make sure you re-familiarize yourself with them.
To thoroughly investigate the function at hand, you have to ask yourself a series of questions, summarized below:
1)
Can I simplify the integral?2)
Does u-substitution work?3)
Is this a trigonometric integral?3)
Is it of a form where trigonometric substitution helps?4)
Will partial fractions help?5) Running out of methods...
will integration-by-parts work?6)
Nothing? Back to the drawing board.
Make sure to start at
question 1, and stop to integrate once you've answered
yes to the question.
Here are some scenarios you may encounter1) Can I simplify the integral? See if you can get the integral to look like one of the integrals in the above table. Make sure to always look for ways to split up an integral into several simpler integrals.
Example 1:\[\displaystyle\int{\dfrac{x-1}{\sqrt{x}}}dx\;\Rightarrow\;\displaystyle\int{\left(x-1\right)\cdot{x}^{-\frac{1}{2}}}dx\;\Rightarrow\;\displaystyle\int{\left({x}^{\frac{1}{2}}-{x}^{-\frac{1}{2}}\right)}dx\]
Example 2:\[\displaystyle\int{\dfrac{\tan\left(x\right)+\cos\left(x\right)}{\sin\left(x\right)}}dx \\=\displaystyle\int{\left(\dfrac{\sin\left(x\right)}{\cos\left(x\right)}+\cos\left(x\right)\right)\cdot\dfrac{1}{\sin\left(x\right)}}dx \\=\displaystyle\int{\left(\dfrac{1}{\cos\left(x\right)}+\dfrac{\cos\left(x\right)}{\sin\left(x\right)}\right)}dx \\=\displaystyle\int{\left(\sec\left(x\right)+\cot\left(x\right)\right)}dx\\ \vdots\]
2) Does u-substitution work? Look for a substitution \(u = g(x)\), where \(g′(x)\) also appears in the integrand (then you know it will become part of the \(du\) after you do the substitution). The end result will hopefully look like one of the integrals listed above.
Example 3:\[\displaystyle\int{\dfrac{1}{x\cdot{\left(\ln\left(x\right)\right)}^{2}}}dx\]
Where:
\[u=\ln(x)\\du=\dfrac{1}{x}dx\\ \therefore dx=xdu\\\displaystyle\int{\dfrac{1}{x\cdot{u}^{2}}}du\\ \vdots\]
Example 4:\[\displaystyle\int{{e}^{{\left(\cos\left(x\right)\right)}^{2}}\sin\left(x\right)\cos\left(x\right)}dx\]
Where:
\[u={\left(\cos\left(x\right)\right)}^{2}\\du=-2\cos\left(x\right)\sin\left(x\right)dx\]
3) Is this a trigonometric integral? If the integrand has powers of \(\cos \left(x\right)\) or \(\sin \left(x\right)\), then follow the rules for dealing with trigonometric integration.
Example 5: (Using half-angle formulas)
\[\displaystyle\int{{\left(\sin\left(3x\right)\right)}^{4}}dx=\displaystyle\int{{\left(\sin\left(3x\right)\right)}^{2}\cdot{\left(\sin\left(3x\right)\right)}^{2}}dx=\displaystyle\int{\dfrac{1}{2}\left(1-\cos\left(6x\right)\right)\cdot\frac{1}{2}\left(1-\cos\left(6x\right)\right)}dx\]
Example 6:\[\displaystyle\int{\dfrac{{\left(\sin\left(x\right)\right)}^{3}}{\sqrt{\cos\left(x\right)}}}dx=\displaystyle\int{\sin\left(x\right)\cdot{\left(\sin\left(x\right)\right)}^{2}{\left(\cos\left(x\right)\right)}^{-\frac{1}{2}}}dx\\=\displaystyle\int{\sin\left(x\right)\left(1-{\left(\cos\left(x\right)\right)}^{2}\right)\left({\left(\cos\left(x\right)\right)}^{-\frac{1}{2}}\right)}dx\]
Where: \(u=\cos x\;\cdots\)
4) Is it of a form where trigonometric substitution helps? If the integrand has a factor of the form \(\sqrt{a^2-x^2}\), \(\sqrt{a^2+x^2}\), or \(\sqrt{x^2+a^2}\), use trigonometric substitution.
Example 7:\[\displaystyle\int{\dfrac{{x}^{3}}{\sqrt{1-{x}^{2}}}}dx\]
Where: \(x=\sin\left(\theta\right)\)
Example 8:\[\displaystyle\int{\dfrac{1}{{x}^{2}\sqrt{{x}^{2}-25}}}dx\]
Where: \(x=5\sec \left(\theta \right)\)
5) Will partial fractions help? If you’re trying to integrate a rational function, check if you can factor the denominator and write the function in terms of its partial fractions.
Example 9:\[\displaystyle\int{\dfrac{6x}{{x}^{2}-2x-3}}dx=\displaystyle\int{\dfrac{6x}{\left(x-3\right)\left(x+1\right)}}dx\;\Rightarrow\;\dfrac{A}{x-3}+\dfrac{B}{x+1}\;\dots\]
6) Running out of methods… will integration by parts work? This is usually used for products of functions. The result is that you end up with a simpler integral, or with the same integral as you started with (“coming full circle”).
Example 10:\[\displaystyle\int{\dfrac{\ln\left(x\right)}{{x}^{7}}}dx=\displaystyle\int{{x}^{-7}\ln\left(x\right)}dx\]
Where:
\[u=\ln\left(x\right)\\dv={x}^{-7}dx\\\vdots\]
Example 11:\[\displaystyle\int{x\cdot{\left(\sec\left(x\right)\right)}^{2}}dx\]
Where:
\[u=x\;\Rightarrow\;du=dx\\dv={\left(\sec\left(x\right)\right)}^{2}dx\;\Rightarrow\;v=\tan\left(x\right)\]
7. Back to the drawing board. If at first you don’t succeed… well, when it comes to integration, you may have to indeed try again a few times. Sometimes integration by parts is the thing to do even though it doesn’t look like it (e.g. using \(dv = dx\)). Sometimes, \(u\)-substitution helps even though \(du\) doesn’t appear in the integral, because it gets it into a form where other methods become clear possibilities. In other words, you may have to combine several techniques, all for one question. Not that bad… just practice.
Example 12: (Integration by Parts)
\[\displaystyle\int{{\left(\tan\left(x\right)\right)}^{-1}}dx\]
Where:
\[u={\left(\tan\left(x\right)\right)}^{-1}\;\Rightarrow\;du=\dfrac{1}{1+{x}^{2}}dx\\dv=dx\;\Rightarrow\;v=x\]
\[uv-\displaystyle\int{v}du=x{\left(\tan\left(x\right)\right)}^{-1}-\displaystyle\int{\dfrac{x}{1+{x}^{2}}}dx\]
Example 13:\[\displaystyle\int{x\sqrt{x+4}}dx\]
Integration by parts:
\[u=x\\dv=\sqrt{x+4}\]
Or by using \(u\)-substitution:
\[u=x+4\;\Rightarrow\;du=dx\\u-4=x\;\Rightarrow\;\displaystyle\int{x\sqrt{x+4}}dx=\displaystyle\int{\left(u-4\right)\sqrt{u}}du\]
Example 14:\[\displaystyle\int{\dfrac{{x}^{3}}{{x}^{2}+2}}dx\]
Where:
\[u={x}^{2}+2\;\Rightarrow\;du=2x \cdot dx\\\dfrac{1}{2}\displaystyle\int{\dfrac{{x}^{2}}{u}}du=\dfrac{1}{2}\displaystyle\int{\dfrac{u-2}{u}}du\\\vdots\]
Or by partial fractions
\[\displaystyle\int{\dfrac{{x}^{3}}{{x}^{2}+2}}dx=\displaystyle\int{\dfrac{x\left({x}^{2}+2\right)-2x}{{x}^{2}+2}}dx=\displaystyle\int{\left(x-\dfrac{2x}{{x}^{2}+2}\right)}dx\]
Or by long division, not shown.
The best way to become an expert at integration is to get lots of practice, so let’s do a bunch of examples 
