Trigonometric Substitution

This is a new method of substitution, and at first, it can be difficult to get used to. The biggest difference from the substitution methods we've used so far (u-substitution and integration by parts) is that rather than defining a new variable explicitly in terms of xx, like u=u= "some expression of xx," we'll define the new variable (θ\theta in this case) implicitly by writing x=x= "some expression of θ\theta."

Just like before, we'll do this so that the integral becomes something we can handle (usually by means of trig integration, as outlined in the previous section), and we'll have to back-substitute in order to have an answer in terms of xx.

Let's illustrate this process with an example.

Evaluate ∫11−x2 dx\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\ dx.

Solution

Look carefully at what we have: 1−x21-x^2. If we let x=sin⁡θx=\sin \theta, this part will become 1−sin⁡2θ1-\sin^2 \theta, which can be replaced with cos⁡2θ\cos^2 \theta, since sin⁡2θ+cos⁡2θ=1.\sin^2 \theta + \cos^2 \theta = 1.

Then we have ∫1cos⁡2θ dx\int \dfrac{1}{\sqrt{\cos^2 \theta}}\ dx

As always, when we do a substitution, we need to account for dxdx: since x=sin⁡θx=\sin \theta, dx=cos⁡θ dθdx = \cos \theta \ d\theta.

That leaves us with ∫1cos⁡2θ cos⁡θ dθ=∫1cos⁡θ cos⁡θ dθ=∫dθ\int \dfrac{1}{\sqrt{\cos^2 \theta}} \ \cos \theta \ d\theta = \int \dfrac{1}{\cos \theta} \ \cos \theta \ d\theta = \int d\theta

This is something we can integrate: ∫dθ=θ+C\int d\theta = \theta + C

All that remains is to substitute back to get the answer in terms of xx: to do this, go back to where we defined θ\theta: if x=sin⁡θx = \sin \theta, θ=sin⁡−1x\theta = \sin^{-1} x.

Thus, ∫11−x2 dx=sin⁡−1x+C\int \dfrac{1}{\sqrt{1-x^2}}\ dx = \ans{\sin^{-1} x + C}

Process

This method is very limited: it applies to exactly three forms. All of these are related to the Pythagorean identity and its variations.

Recall that sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 and if we divide each term by cos⁡2θ\cos^2 \theta, we get tan⁡2θ+1=sec⁡2θ.\tan^2 \theta + 1 = \sec^2 \theta.

We've already seen the first form: the problem involved 1−x2\sqrt{1-x^2}, so we let x=sin⁡θx=\sin \theta so that the first identity helped us simplify the integral. We could also use x=cos⁡θx=\cos \theta in this case, but for simplicity we'll stick with the sine function.

Case 1: x=asin⁡θx = a \sin \theta

When the integral involves a2−x2\sqrt{a^2-x^2}, let x=asin⁡θx=a \sin \theta (in the example above, a=1a=1).

Then dx=acos⁡θ dθdx = a\cos \theta \ d\theta and a2−x2=a2−a2sin⁡2θ=a1−sin⁡2θ=acos⁡2θ=acos⁡θ\begin{aligned} \sqrt{a^2-x^2} &= \sqrt{a^2 - a^2\sin^2 \theta}\ &= a\sqrt{1-\sin^2 \theta}\ &= a\sqrt{\cos^2 \theta}\ &= a \cos \theta \end{aligned}

Evaluate ∫4−x2 dx\displaystyle\int \sqrt{4-x^2} \ dx.

Solution

In this case, a=2a=2 (a2=4a^2=4), so we'll let x=2sin⁡θx=2 \sin \theta.

Then dx=2cos⁡θ dθdx = 2 \cos \theta \ d\theta and 4−x2=2cos⁡θ\sqrt{4-x^2} = 2 \cos \theta

Making the substitution: ∫4−x2 dx=∫2cos⁡θ⋅2cos⁡θ dθ\int \sqrt{4-x^2} \ dx = \int 2 \cos \theta \cdot 2 \cos \theta \ d\theta

To integrate this, recall what we did in the previous section: to integrate cos⁡2θ\cos^2 \theta, we'll need to use the half-angle identity: ∫4cos⁡2θ dθ=∫4(12+12cos⁡(2θ)) dθ=∫2+2cos⁡(2θ) dθ=2θ+sin⁡(2θ)+C\begin{aligned} \int 4 \cos^2 \theta \ d\theta &= \int 4 \left(\dfrac{1}{2} + \dfrac{1}{2}\cos(2\theta)\right)\ d\theta\ &= \int 2 + 2\cos (2\theta) \ d\theta\ &= 2\theta + \sin(2\theta) + C \end{aligned}

Get the answer in terms of xx

This part is a bit tricky. First, we need another identity that we haven't used before: sin⁡(2θ)=2sin⁡θcos⁡θ\sin (2\theta) = 2 \sin \theta \cos \theta

This makes our answer look like 2θ+2sin⁡θcos⁡θ+C2\theta + 2\sin \theta \cos \theta + C

Now finally, we need to figure out what θ\theta, sin⁡θ\sin \theta, and cos⁡θ\cos \theta are in terms of xx.

The first we've already seen: if x=2sin⁡θx=2\sin \theta, we can solve for θ\theta: θ=sin⁡−1x2\theta = \sin^{-1} \dfrac{x}{2}

Also, from this we can see that sin⁡θ=x2.\sin \theta = \dfrac{x}{2}.

The last part is to find cos⁡θ\cos \theta. There are two ways to do this:

Algebraic Method

Look back at what we have at the beginning of the problem: 4−x2=2cos⁡θ⟶cos⁡θ=4−x22\sqrt{4-x^2} = 2 \cos \theta \longrightarrow \cos \theta = \dfrac{\sqrt{4-x^2}}{2}

Geometric Method

Since sin⁡θ=x2\sin \theta = \dfrac{x}{2}, we can draw the following right triangle and use it to find cos⁡θ\cos \theta:

Right triangle to find cosine

We get the same answer from this method: cos⁡θ=4−x22\cos \theta = \dfrac{\sqrt{4-x^2}}{2}

Final Answer

Putting it all together: ∫4−x2 dx=∫4cos⁡2θ dθ=2θ+sin⁡(2θ)+C=2θ+2sin⁡θcos⁡θ+C=2sin⁡−1x2+2(x2)(4−x22)+C=2sin⁡−1x2+12x4−x2+C\begin{aligned} \int \sqrt{4-x^2}\ dx &= \int 4 \cos^2 \theta \ d\theta\ &= 2\theta + \sin (2\theta) + C\ &= 2\theta + 2 \sin \theta \cos \theta + C\ &= 2\sin^{-1} \dfrac{x}{2} + 2\left(\dfrac{x}{2}\right)\left(\dfrac{\sqrt{4-x^2}}{2}\right) + C\ &= \ans{2\sin^{-1} \dfrac{x}{2} + \dfrac{1}{2}x\sqrt{4-x^2} + C} \end{aligned}

Case 2: x=atan⁡θx = a \tan \theta

When the integral involves a2+x2\sqrt{a^2 + x^2}, let x=atan⁡θx= a \tan \theta. Then dx=asec⁡2θ dθdx = a \sec^2 \theta \ d\theta and a2+x2=a2+a2tan⁡2θ=a1+tan⁡2θ=asec⁡2θ=asec⁡θ\begin{aligned} \sqrt{a^2 + x^2} &= \sqrt{a^2 + a^2 \tan^2 \theta}\ &= a\sqrt{1+\tan^2 \theta}\ &= a\sqrt{\sec^2 \theta}\ &= a \sec \theta \end{aligned}

Evaluate ∫x3x2+9 dx\displaystyle\int \dfrac{x^3}{\sqrt{x^2+9}}\ dx.

Solution

Here a=3a = 3, so let x=3tan⁡θx = 3 \tan \theta. Then dx=3sec⁡2θ dθdx = 3 \sec^2 \theta\ d\theta and x2+9=3sec⁡θ.\sqrt{x^2+9} = 3 \sec \theta.

Substitute and simplify: ∫x3x2+9 dx=∫(3tan⁡θ)33sec⁡θ⋅3sec⁡2θ dθ=∫27tan⁡3θsec⁡θ dθ\begin{aligned} \int \dfrac{x^3}{\sqrt{x^2+9}}\ dx &= \int \dfrac{(3 \tan \theta)^3}{3 \sec \theta} \cdot 3 \sec^2 \theta \ d\theta\ &= \int 27 \tan^3 \theta \sec \theta \ d\theta \end{aligned}

Then look back again at what we did in the previous section: since the power of tangent is odd, we'll let u=sec⁡θu=\sec \theta, so that du=sec⁡θtan⁡θ dθdu = \sec \theta \tan \theta \ d\theta. Then we'll use the identity tan⁡2θ=sec⁡2θ−1\tan^2 \theta = \sec^2 \theta - 1 to rewrite the remainder in terms of sec⁡θ\sec \theta: ∫27tan⁡3θsec⁡θ dθ=∫27tan⁡2θ sec⁡θtan⁡θ dθ=∫27(sec⁡2θ−1)sec⁡θtan⁡θ dθ=∫27u2−27 du=9u3−27u+C=9sec⁡3θ−27sec⁡θ+C\begin{aligned} \int 27 \tan^3 \theta \sec \theta \ d\theta &= \int 27 \tan^2 \theta \ \sec \theta \tan \theta \ d\theta\ &= \int 27 (\sec^2 \theta - 1) \sec \theta \tan \theta \ d\theta\ &= \int 27u^2 - 27 \ du\ &= 9u^3 - 27u + C\ &= 9\sec^3 \theta - 27 \sec \theta + C \end{aligned}

Finally, to get the answer in terms of xx, notice that we found that x2+9=3sec⁡θ\sqrt{x^2+9} = 3 \sec \theta so sec⁡θ=x2+93\sec \theta = \dfrac{\sqrt{x^2 + 9}}{3}.

The answer, then, is ∫x3x2+9 dx=9(x2+93)3−27(x2+93)+C\int \dfrac{x^3}{\sqrt{x^2+9}}\ dx = 9\left(\dfrac{\sqrt{x^2 + 9}}{3}\right)^3 - 27 \left(\dfrac{\sqrt{x^2 + 9}}{3}\right) + C

After a bit of simplification, we get 13(x2+9)3−9x2+9+C\ans{\dfrac{1}{3}(\sqrt{x^2+9})^3 - 9\sqrt{x^2+9} + C}

Case 3: x=asec⁡θx= a \sec \theta

When the integral involves x2−a2\sqrt{x^2-a^2}, let x=asec⁡θx=a\sec \theta. Then dx=asec⁡θtan⁡θ dθdx = a \sec \theta \tan \theta\ d\theta and x2−a2=a2sec⁡2θ−a2=asec⁡2θ−1=atan2θ=atan⁡θ\begin{aligned} \sqrt{x^2-a^2} &= \sqrt{a^2 \sec^2 \theta - a^2}\ &= a\sqrt{\sec^2 \theta - 1}\ &= a \sqrt{tan^2 \theta}\ &= a \tan \theta \end{aligned}

Evaluate ∫x2−1x dx\displaystyle\int \dfrac{\sqrt{x^2-1}}{x}\ dx.

Solution

Since this fits the third case with a=1a=1, let x=sec⁡θx = \sec \theta. Thus dx=sec⁡θtan⁡θ dθdx = \sec \theta \tan \theta\ d\theta and x2−1=tan⁡θ.\sqrt{x^2-1} = \tan \theta.

Substitute: ∫x2−1x dx=∫tan⁡θsec⁡θ⋅sec⁡θtan⁡θ dθ=∫tan⁡2θ dθ\begin{aligned} \int \dfrac{\sqrt{x^2-1}}{x}\ dx &= \int \dfrac{\tan \theta}{\sec \theta} \cdot \sec \theta \tan \theta \ d\theta\ &= \int \tan^2 \theta \ d\theta \end{aligned}

To integrate this, replace tan⁡2θ\tan^2 \theta with sec⁡2θ−1\sec^2 \theta - 1 and remember that ∫sec⁡2θ dθ=tan⁡θ+C\int \sec^2 \theta \ d\theta = \tan \theta + C Thus ∫x2−1x dx=∫tan⁡2θ dθ=∫sec⁡2θ−1 dθ=tan⁡θ−θ+C\begin{aligned} \int \dfrac{\sqrt{x^2-1}}{x}\ dx &= \int \tan^2 \theta \ d\theta\ &= \int \sec^2 \theta - 1\ d\theta\ &= \tan \theta - \theta + C \end{aligned}

Finally, look back at the beginning of the problem, and you'll find that tan⁡θ=x2−1 and θ=sec⁡−1x\tan \theta = \sqrt{x^2-1} \textrm{ and } \theta = \sec^{-1} x so the final answer is ∫x2−1x dx=x2−1−sec⁡−1x+C\int \dfrac{\sqrt{x^2-1}}{x}\ dx = \ans{\sqrt{x^2-1} - \sec^{-1} x + C}

  1. ∫x39−x2 dx\displaystyle\int x^3 \sqrt{9-x^2}\ dx

    2. Let x=3sin⁡θx=3 \sin \theta. Then dx=3cos⁡θ dθdx=3\cos \theta \ d\theta and 9−x2=3cos⁡θ\sqrt{9-x^2} = 3\cos \theta. ∫x39−x2 dx=∫27sin⁡3θ⋅3cos⁡θ⋅3cos⁡θ dθ=∫243sin⁡3θcos⁡2θ dθ=∫243sin⁡2θcos⁡2θsin⁡θ dθ=∫243(1−cos⁡2θ)cos⁡2θsin⁡θ dθ\begin{aligned} \int x^3 \sqrt{9-x^2}\ dx &= \int 27 \sin^3 \theta \cdot 3 \cos \theta \cdot 3 \cos \theta \ d\theta\ &= \int 243 \sin^3 \theta \cos^2 \theta \ d\theta\ &= \int 243 \sin^2 \theta \cos^2 \theta \sin \theta \ d\theta\ &= \int 243 (1 - \cos^2 \theta) \cos^2 \theta \sin \theta \ d\theta \end{aligned} Then let u=cos⁡θu=\cos \theta, so that −du=sin⁡θ dθ-du = \sin \theta\ d\theta. ∫243(1−cos⁡2θ)cos⁡2θsin⁡θ dθ=∫−243(1−u2)u2 du=∫−243u2+243u4 du=−81u3+2435u5+C=−81cos⁡3θ+2435cos⁡5θ+C=−3(9−x2)3+15(9−x2)5+C\begin{aligned} \int 243 (1 - \cos^2 \theta) \cos^2 \theta \sin \theta \ d\theta &= \int -243 (1-u^2) u^2 \ du\ &= \int -243u^2 + 243u^4 \ du\ &= -81u^3 + \dfrac{243}{5}u^5 + C\ &= -81\cos^3 \theta + \dfrac{243}{5}\cos^5 \theta + C\ &= \ans{-3(\sqrt{9-x^2})^3 + \dfrac{1}{5}(\sqrt{9-x^2})^5 + C} \end{aligned}

  2. ∫x3x2+100 dx\displaystyle\int \dfrac{x^3}{\sqrt{x^2+100}}\ dx

    3. Let x=10tan⁡θx=10 \tan \theta. Then dx=10sec⁡2θ dθdx=10\sec^2 \theta \ d\theta and x2+100=10sec⁡θ\sqrt{x^2+100} = 10\sec \theta. ∫x3x2+100 dx=∫1000tan⁡3θ10sec⁡θ⋅10sec⁡2θ dθ=∫1000tan⁡3θsec⁡θ dθ=∫1000tan⁡2θsec⁡θtan⁡θ dθ=∫1000(sec⁡2θ−1)sec⁡θtan⁡θ dθ\begin{aligned} \int \dfrac{x^3}{\sqrt{x^2+100}}\ dx &= \int \dfrac{1000 \tan^3 \theta}{10 \sec \theta} \cdot 10\sec^2 \theta \ d\theta\ &= \int 1000 \tan^3 \theta \sec \theta \ d\theta\ &= \int 1000 \tan^2 \theta \sec \theta \tan \theta \ d\theta\ &= \int 1000 (\sec^2 \theta - 1) \sec \theta \tan \theta \ d\theta \end{aligned} Then let u=sec⁡θu=\sec \theta, so that du=sec⁡θtan⁡θ dθdu = \sec \theta \tan \theta\ d\theta. ∫1000(sec⁡2θ−1)sec⁡θtan⁡θ dθ=∫1000(u2−1) du=10003u3−1000u+C=10003sec⁡3θ−1000sec⁡θ+C=13(x2+100)3−100x2+100+C\begin{aligned} \int 1000 (\sec^2 \theta - 1) \sec \theta \tan \theta \ d\theta &= \int 1000 (u^2-1) \ du\ &= \dfrac{1000}{3}u^3 - 1000u + C\ &= \dfrac{1000}{3}\sec^3 \theta - 1000 \sec \theta + C\ &= \ans{\dfrac{1}{3}(\sqrt{x^2+100})^3 - 100 \sqrt{x^2+100} + C} \end{aligned}

  3. ∫x316−x2 dx\displaystyle\int \dfrac{x^3}{\sqrt{16-x^2}}\ dx

    4. Let x=4sin⁡θx=4 \sin \theta. Then dx=4cos⁡θ dθdx=4\cos \theta \ d\theta and 16−x2=4cos⁡θ\sqrt{16-x^2} = 4\cos \theta. ∫x316−x2 dx=∫64sin⁡3θ4cos⁡θ⋅4cos⁡θ dθ=∫64sin⁡3θ dθ=∫64sin⁡2θsin⁡θ dθ=∫64(1−cos⁡2θ)sin⁡θ dθ\begin{aligned} \int \dfrac{x^3}{\sqrt{16-x^2}}\ dx &= \int \dfrac{64 \sin^3 \theta}{4 \cos \theta} \cdot 4 \cos \theta \ d\theta\ &= \int 64 \sin^3 \theta \ d\theta\ &= \int 64 \sin^2 \theta \sin \theta \ d\theta\ &= \int 64 (1 - \cos^2 \theta) \sin \theta \ d\theta \end{aligned} Then let u=cos⁡θu=\cos \theta, so that −du=sin⁡θ dθ-du = \sin \theta\ d\theta. ∫64(1−cos⁡2θ)sin⁡θ dθ=∫−64(1−u2) du=∫−64+64u2 du=−64u+643u3+C=−64cos⁡θ+643cos⁡3θ+C=−1616−x2+13(16−x2)3+C\begin{aligned} \int 64 (1 - \cos^2 \theta) \sin \theta \ d\theta &= \int -64 (1-u^2) \ du\ &= \int -64 + 64u^2 \ du\ &= -64u + \dfrac{64}{3}u^3 + C\ &= -64\cos \theta + \dfrac{64}{3}\cos^3 \theta + C\ &= \ans{-16\sqrt{16-x^2} + \dfrac{1}{3}(\sqrt{16-x^2})^3 + C} \end{aligned}

  4. ∫xx2+4 dx\displaystyle\int x \sqrt{x^2+4}\ dx

    5. Let x=2tan⁡θx=2 \tan \theta. Then dx=2sec⁡2θ dθdx=2\sec^2 \theta \ d\theta and x2+4=2sec⁡θ\sqrt{x^2+4} = 2\sec \theta. ∫xx2+4 dx=∫2tan⁡θ⋅2sec⁡θ⋅2sec⁡2θ dθ=∫8tan⁡θsec⁡3θ dθ=∫8sec⁡2θsec⁡θtan⁡θ dθ\begin{aligned} \int x \sqrt{x^2+4}\ dx &= \int 2\tan \theta \cdot 2 \sec \theta \cdot 2 \sec^2 \theta \ d\theta\ &= \int 8 \tan \theta \sec^3 \theta \ d\theta\ &= \int 8 \sec^2 \theta \sec \theta \tan \theta \ d\theta\ \end{aligned} Then let u=sec⁡θu=\sec \theta, so that du=sec⁡θtan⁡θ dθdu = \sec \theta \tan \theta\ d\theta. ∫8sec⁡2θsec⁡θtan⁡θ dθ=∫8u2 du=83u3+C=83sec⁡3θ+C=13(x2+4)3+C\begin{aligned} \int 8 \sec^2 \theta \sec \theta \tan \theta \ d\theta &= \int 8u^2 \ du\ &= \dfrac{8}{3}u^3 + C\ &= \dfrac{8}{3}\sec^3 \theta + C\ &= \ans{\dfrac{1}{3}(\sqrt{x^2+4})^3 + C} \end{aligned}