Using the unique algebra of trig. functions to solve integrals

In another section we looked at integrals containing only powers of trigonometric functions, such as

$$ \begin{align} &\int \text{sin}^3(x) \, dx \\[5pt] &\int \text{sin}^2(x) \, \text{tan}(x) \, dx \\[5pt] &\int \text{cos}^4(x) \, \text{sin}^2(x) \, dx \end{align}$$

We developed a couple of techniques, relying on key trigonometric identities, to solve them. These are an important class of integrals in certain fields of mathematics, particularly in the field of Fourier series and the Fourier transform.

In this section, we'll look at quite a different set of integrands, functions that look like this:

How it works

Let's start with the first radical expression in the list above, $\sqrt{a^2 - x^2}$. We begin by introducing a dummy variable, $t$, and making the substitution $x = a \, \text{sin}(t)$. We'll get rid of the $t$ later when we're done with it, so hang in there. Here's the process:

$$ \begin{align} \sqrt{a^2-x^2} &= \sqrt{a^2 - a^2 \text{sin}^2(t)} \\[5pt] &= \sqrt{a^2 \, (1 - \text{sin}^2(t))} \\[5pt] &= \sqrt{a^2 \, \text{cos}^2(t)} \\[5pt] &= a \, \text{cos}(t) \end{align}$$

Table of common trigonometric substitutions

Example 1   $\int \sqrt{81 - x^2} \, dx$

To begin, consult the table above and make the substitution $x = a \, \text{sin}(t)$, where $a = 9$ (the square root of 81):

$$ \begin{align} \text{let } \; x &= 9 \, \text{sin}(t) \\[5pt] \text{then } \; x^2 &= 81 \, \text{sin}^2(t) \\[5pt] \text{and } \; dx &= 9 \, \text{cos}(t) \, dt \end{align}$$

The integral is

$$\int \sqrt{81 - x^2} \, dx$$

The integrand under the radical is $9^2 (1 - \text{sin}^2(t))$, which is a perfect square:

$$ \begin{align} 81 - 81 \text{sin}^2(t) &= 81[1 - \text{sin}^2(t)] \\[5pt] &= 81 \text{cos}^2(t) \end{align}$$

That last step uses the Pythagorean identity, $\text{sin}^2(t) + \text{cos}^2(t) = 1$. Now the integrand, including $dx$, can be written:

$$\int \sqrt{81 \, \text{cos}^2(t)} \cdot 9 \text{cos}(t) \, dt$$

The substitution allows us to take the square root:

$$\int 9 \text{cos}(t) \cdot 9 \text{cos}(t) \, dt$$

Now move the constant outside of the integral and use our co\text{sin}e power-reduction formula to change the $\text{cos}^2(t)$ term into something we can integrate, just as in the examples above. Integrate the $\text{cos}(2t)$ integrand with simple u-substitution.

$$81 \int \text{cos}^2(t) \, dt = \frac{81}{2} \int 1 + \text{cos}(2t) \, dt$$

Example 2   $\int \sqrt{4 + 9x^2} \, dx$

This integral looks like the form

$$\int \sqrt{a^2 + b^2} \, dx$$

except for the number (9) in front of the $x^2$. We can remove that temporarily by dividing by the square root of 9, and multiplying outside the integrand by 3. We'll have to keep that 3 around, but that's OK:

$$3 \int \frac{\sqrt{4 + 9x^2}}{\sqrt{9}} \, dx$$

The new integral now has the form we'd like:

$$3 \int \sqrt{\frac{4}{9} + x^2} \, dx$$

Consulting the table above, we see that this substitution works:

$$ \begin{align} \text{let } \; x &= \frac{2}{3} \, \text{tan}(t), \\[5pt] \text{then } \; x^2 &= \frac{4}{9} \, \text{tan}^2(t) \\[5pt] \text{and } \; dx &= \frac{2}{3} \, \text{sec}^2(t) \, dt \end{align}$$

The radical in the integrand now reduced to

$$ \begin{align} \sqrt{\frac{4}{9} + x^2} &= \sqrt{\frac{4}{9} + \frac{4}{9} \text{tan}^2(t)} \\[5pt] &= \sqrt{\frac{4}{9}[1 + \text{tan}^2(t)]} \\[5pt] &= \sqrt{\frac{4}{9} \text{sec}^2(t)} \\[5pt] &= \frac{2}{3} \text{sec}(t) \end{align}$$

Multiplying through by $\text{sec}(t)$ on the right side gives us two integrals. Notice that the integral of $\text{sec}^3(t)$ reappears, but with opposite sign, so we can move it to the left side by addition (arrow).

Now the integral of $\text{sec}(t)$ was worked out elsewhere, it is $ln|\text{sec}(t) + \text{tan}(t)|$:

$$2 \int \text{sec}^3(t) \, dt = \text{sec}(t) \cdot \text{tan}(t) - ln|\text{sec}(t) + \text{tan}(t)| + C$$

We can finally solve for the original integral, remembering to bring our factor of 4/3 back in:

$$\int \text{sec}^3(t) \, dt = \frac{4}{3}\frac{1}{2}\left[ \text{sec}(t) \cdot \text{tan}(t) - ln|\text{sec}(t) + \text{tan}(t) + C \right]$$

The solution, in terms of the dummy variable $>t$ is:

$$= \frac{2}{3} \left[ \text{sec}(t) \cdot \text{tan}(t) = ln|\text{sec}(t) + \text{tan}(t)| + C\right]$$

Now we can construct a triangle based on our initial assumption, $\text{tan}(t) = \frac{3x}{2}$

The hypotenuse was found using the Pythagorean theorem. We can now use the triangle to find $\text{tan}(t)$ and $\text{sec}(t)$ in terms of the original variable $x$:

$$= \frac{\sqrt{4 + 9x^2}}{2} \frac{3x}{2} - ln \left| \frac{\sqrt{4 + 9x^2}}{2} + \frac{3x}{2} \right| + C$$

OK, so that was a long slog, but hopefully you can see that there's just no other way to do that integral on paper. Our only other choice would have been to estimate a definite integral by numerical integration.

Example 3:   $\int \frac{e^x}{7 + e^{2x}} \, dx$

The first thing to do is make a substitution for the ex terms. This one fits nicely:

$$\text{Let } u = e^x, \; \text{ then } u^2 = e^{2x}$$

(remember the laws of logs: $(e^x)^2 = e^{2x}$). We also need the differential in terms of $u$:

$$\text{and } du = e^x \, dx$$

The new integral is

$$\int \frac{du}{7 + u^2}$$

Now this has a form for which we can use the $x = a·\text{tan}(t)$ substitution:

$$ \begin{align} \text{Let } u = \sqrt{7} \, \text{tan}(t) \\[5pt] du = \sqrt{7} \, \text{sec}^2(t) \end{align}$$

The denominator gets substituted like this:

$$ \begin{align} u^2 &= 7 \, \text{tan}^2(t) \\[5pt] 7 + u^2 &= 7 + 7 \, \text{tan}^2(t) \\[5pt] &= 7(1 + \text{tan}^2(t)) \\[5pt] &= 7 \, \text{sec}^2(t) \end{align}$$

Practice problems

Try doing these integrals by trig. substitution. Some of them are tricky. Try u\text{sin}g the solutions as hints to get you going when you stall. You'll learn more that way.

1. $$\int \frac{1}{\sqrt{x^2 - 25}} \, dx$$

   Solution

   $$ \begin{align} Let \; x &= 5 \text{sec}(t) \\[5pt] x^2 &= 25 \text{sec}^2(t) \\[5pt] x^2 - 25 &= 25(\text{sec}^2(t) - 1) = 25 \text{tan}^2(t)\\[5pt] \sqrt{x^2 - 25} &= 5 \text{tan}(t) \; and \\[5pt] dx &= 5 \text{sec}(t) \text{tan}(t) \, dt \\[5pt] \longrightarrow \int \frac{5 \text{sec}(t) \text{tan}(t)}{5 \text{tan}(t)} &= \int \text{sec}(t) \, dt \\[5pt] &= ln|\text{sec}(t) + \text{tan}(t)| + C \end{align}$$

   $$Now \; \text{sec}(t) = \frac{x}{5} \text{ gives us this triangle:}$$

   

   $$ \begin{align} \int \frac{1}{\sqrt{x^2 - 25}} \, dx &= ln \left| \frac{x}{5} + \frac{\sqrt{x^2 - 25}}{5} \right| + C \\[5pt] &= ln \left| \frac{x + \sqrt{x^2 - 25}}{5} \right| + C \end{align}$$




2. $$\int \frac{x}{\sqrt{16 - x^2}} \, dx$$

   Solution

   $$ \begin{align} Let \; x &= 4 \text{sin}(t) \\[5pt] x^2 &= 16 \text{sin}^2(t) \\[5pt] 16 - x^2 &= 16(1 - \text{sin}^2(t)) = 16 \text{cos}^2(t)\\[5pt] \sqrt{16 - x^2} &= 4 \text{cos}(t) \; and \\[5pt] dx &= 4 \text{cos}(t) \, dt \\[5pt] \longrightarrow \int \frac{4 \text{cos}(t)}{4 \text{cos}(t)} \, dt &= \int \, dt = t + C \end{align}$$

   $$Now \; \text{sin}(t) = \frac{x}{4} \; so \; \; t = \text{sin}^{-1}\frac{x}{4}$$

   $$= \int \frac{x}{\sqrt{16 - x^2}} \, dx = \text{sin}^{-1} \left( \frac{x}{4} \right) + C$$




3. $$\int \frac{x}{\sqrt{x^2 + 1}} \, dx$$

   Solution

   $$ \begin{align} Let \; x &= 1\cdot \text{tan}(t) \\[5pt] x^2 &= \text{tan}^2(t) \\[5pt] x^2 + 1 &= \text{tan}^2(t) + 1 = \text{sec}^2(t)\\[5pt] \sqrt{x^2 + 1} &= \text{sec}(t) \; and \\[5pt] dx &= \text{sec}^2(t) \, dt \\[5pt] \int \frac{x}{\sqrt{x^2 + 1}} \, dx &\longrightarrow \int \frac{\text{tan}(t) \text{sec}^2(t)}{\text{sec}(t)}\, dt \\[5pt] &= \int \text{sec}(t)\text{tan}(t) \, dt \\[5pt] &= \text{sec}(t) + C \end{align}$$

   $$\text{Now} \; \text{tan}(t) = \frac{x}{1} \text{ gives us this right triangle:}$$

   

   $$\int \frac{1}{\sqrt{x^2 + 1}} \, dx = \sqrt{x^2 + 1} + C$$




4. $$\int \frac{x^2}{\sqrt{x^2 + 9}} \, dx$$

   Solution

   $$\int \frac{x^2}{\sqrt{x^2 + 9}} \, dx$$

   $$ \begin{align} Let \; x &= 3 \text{tan}(t) \\[5pt] x^2 &= 9 \text{tan}^2(t) \\[5pt] x^2 + 9 &= 9(\text{tan}^2(t) + 1) = 9 \text{sec}^2(t) \\[5pt] \sqrt{x^2 + 9} &= 3 \text{sec}(t), \; and \\[5pt] x^2 &= 9 \text{tan}^2(t), \; and \\[5pt] dx &= 3 \text{sec}^2(t) \, dt \\[5pt] \int \frac{x^2}{\sqrt{x^2 + 9}} \, dx &\longrightarrow \int \frac{9 \text{tan}^2(t) \cdot 3 \text{sec}^2(t) \, dt}{3 \text{sec}(t)} \\ &= 9 \int \text{tan}^2(t) \text{sec}(t) \, dt \end{align}$$

   Now this integral can be done using all of the rest of our integration skill ... hold on: First use the identity   $\text{tan}^2(x) = \text{sec}^2(x) - 1$, so our integral breaks into two parts — and let's leave the 9 off for now to simplify things; we'll put it back later.

   $$ \int \text{sec}^3(t)\, dt - \int \text{sec}(t) \, dt$$

   The integral on the right is   $ \int \text{sec}(t) \, dt = ln|\text{sec}(t) + \text{tan}(t)| + C$

   The integral on the left can be done by parts, letting   $u = \text{sec}(t)$, and   $dv = \text{sec}^2(t) \, dt$. Then   $du = \text{sec}(t) \text{tan}(t) \, dt$ and   $v = \text{tan}(t)$, so for the left-side integral, we have:

   $$\int \text{sec}^3(x) \, dt = \text{sec}(t) \text{tan}(t) - \int \text{tan}^2(t) \text{sec}(t) \, dt$$

   Now using the same identity as above, we have:

   $$ \begin{align} \int \text{sec}^3(t) \, dt &= \text{sec}(t) \text{tan}(t) - \int (\text{sec}^2(t) - 1) \text{sec}(t) \, dt \\[5pt] &= \text{sec}(t) \text{tan}(t) - \left[ \int \text{sec}^3(t) \, dt - \int \text{sec}(t) \, dt \right] \end{align}$$

   So we have:

   $$2\int \text{sec}^3(x) \, dt = \text{sec}(t) \text{tan}(t) - ln|\text{sec}(t) + \text{tan}(t)|$$

   $$\int \text{sec}^3(t)\, dt = \frac{1}{2} \text{sec}(t) \text{tan}(t) + \frac{1}{2} ln|\text{sec}(t) + \text{tan}(t)|$$

   Now putting it all together, we get:

   $$\int \frac{x^2}{\sqrt{x^2 + 9}} \, dx = \frac{9}{2} \text{sec}(t) \text{tan}(t) - \frac{9}{2} ln|\text{sec}(t) + \text{tan}(t)|$$

   Now our original trig substitution was $\text{tan}(t) = \frac{x}{3}$, which gives us this right triangle:

   

   We can use that triangle to get the final integral:

   $$ \begin{align} \int \frac{x^2}{\sqrt{x^2 + 9}} \, dx &= \frac{9}{2} \frac{x \sqrt{x^2 + 9}}{9} - \frac{9}{2} ln \left| \frac{\sqrt{x^2 + 9}}{3} + \frac{x}{3}\right| \\[5pt] &= \frac{1}{2} x \sqrt{x^2 + 9} - \frac{9}{2} ln\left|\frac{\sqrt{x^2 + 9}}{3} + \frac{x}{3}\right| \end{align}$$