Inverse substitution: trigonometric substitution

In substitution (week 12) you defined a new variable by saying $u=g(x)$. In inverse substitution you do the opposite: you write $x$ as a function of a new variable, $x=g(\theta)$. Why? Because for roots like $\sqrt{a^2-x^2}$, putting $x=a\sin\theta$ gives $\sqrt{a^2\cos^2\theta}=a\cos\theta$ and the root disappears. This is using the trigonometric identities ($1-\sin^2=\cos^2$, $1+\tan^2=\sec^2$, $\sec^2-1=\tan^2$) in integrals. There is usually one on the final exams. Keep in mind:

1. The form inside the root determines the substitution. There are three types, memorize them.
2. Convert $dx$ too. If $x=a\sin\theta$, then $dx=a\cos\theta\,d\theta$.
3. Draw a triangle to convert the result back to $x$.
4. Try a simple substitution first. If the derivative of the radicand is outside the root, trigonometry is unnecessary.

The three basic substitutions

Because you choose $\theta$ from the range of the inverse trigonometric function ($[-\tfrac\pi2,\tfrac\pi2]$ for sine), $\cos\theta\ge0$, and the square root comes out without an absolute value.

Sine substitution: $\sqrt{a^2-x^2}$

Example 1. $\displaystyle\int\sqrt{9-x^2}\,dx$.

x = 3 sin θ,   dx = 3 cos θ dθ,   sqrt(9 - x^2) = 3 cos θ
∫ 3 cos θ · 3 cos θ dθ = 9 ∫ cos^2 θ dθ
                       = 9 (θ/2 + sin 2θ / 4)
                       = (9/2) θ + (9/2) sin θ cos θ

Now go back to $x$. Draw the right triangle with $\sin\theta=\tfrac x3$: opposite side $x$, hypotenuse $3$, adjacent side $\sqrt{9-x^2}$. From it $\cos\theta=\tfrac{\sqrt{9-x^2}}{3}$ and $\theta=\arcsin\tfrac x3$.

$\int\sqrt{9-x^2}\,dx=\frac92\arcsin\frac x3+\frac x2\sqrt{9-x^2}+C$

Check: $\int_0^3\sqrt{9-x^2}\,dx$ is the area of a quarter circle, $\tfrac{9\pi}{4}$. From the formula: $\tfrac92\cdot\tfrac\pi2+0-0=\tfrac{9\pi}{4}$. It matches.

PITFALL · leaving $\sin2\theta$ unconverted. If $\theta$ or $\sin2\theta$ remains in the answer, it counts as incomplete. First write $\sin2\theta=2\sin\theta\cos\theta$, then go to $x$ using the triangle.

Tangent substitution: $\sqrt{a^2+x^2}$

Example 2. $\displaystyle\int\frac{dx}{\sqrt{x^2+4}}$.

$x=2\tan\theta$, $dx=2\sec^2\theta\,d\theta$, $\sqrt{x^2+4}=2\sec\theta$.

$\int\frac{2\sec^2\theta}{2\sec\theta}d\theta=\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C$

Triangle: $\tan\theta=\tfrac x2$ (opposite $x$, adjacent $2$), hypotenuse $\sqrt{x^2+4}$, $\sec\theta=\tfrac{\sqrt{x^2+4}}{2}$.

$\ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+C=\ln\left(x+\sqrt{x^2+4}\right)+C'$

$-\ln2$ is absorbed into the constant. In week 7 you saw that the derivative of $\ln(x+\sqrt{x^2+1})$ is $\tfrac{1}{\sqrt{x^2+1}}$; this is its general form.

Secant substitution: $\sqrt{x^2-a^2}$

Example 3. $\displaystyle\int\frac{dx}{x^2\sqrt{x^2-1}}$, $x>1$.

$x=\sec\theta$, $dx=\sec\theta\tan\theta\,d\theta$, $\sqrt{x^2-1}=\tan\theta$.

$\int\frac{\sec\theta\tan\theta}{\sec^2\theta\tan\theta}d\theta=\int\cos\theta\,d\theta=\sin\theta+C$

Triangle: $\sec\theta=\tfrac x1$, so the hypotenuse is $x$, adjacent $1$, opposite $\sqrt{x^2-1}$. $\sin\theta=\dfrac{\sqrt{x^2-1}}{x}$.

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

Inverse substitution in a definite integral

In a definite integral you don’t have to go back to $x$: if you convert the limits to $\theta$, you finish without drawing a triangle. For $x=a\sin\theta$, $\theta=\arcsin\tfrac xa$; for $x=a\tan\theta$, $\theta=\arctan\tfrac xa$. Convert each limit with this formula.

Example 3b. $\displaystyle\int_0^{\sqrt3}\frac{dx}{(1+x^2)^2}$.

x = tan θ,  dx = sec^2 θ dθ,  (1 + x^2)^2 = sec^4 θ
x = 0   -> θ = 0          x = √3 -> θ = arctan √3 = π/3
∫_0^(π/3) sec^2 θ / sec^4 θ dθ = ∫_0^(π/3) cos^2 θ dθ
  = [θ/2 + sin 2θ / 4]_0^(π/3) = π/6 + (√3/2)/4 = π/6 + √3/8 ≈ 0.740

There is no root here but there is a power of $1+x^2$; the tangent substitution works again. The rule is not limited to roots: think of the substitution whenever the expression $a^2\pm x^2$ or $x^2-a^2$, or its power, makes the integrand hard.

Why does it work, and why do we go back?

In inverse substitution you are really changing the variable with the function $\theta=\arcsin\tfrac xa$. Since this function is one-to-one on the chosen interval, the substitution can be undone and the integral does not change. Knowing the range of $\theta$ ($[-\tfrac\pi2,\tfrac\pi2]$, and $[0,\tfrac\pi2)$ for secant) determines the sign of the $\cos\theta$ or $\tan\theta$ that comes out of the root. That is why Example 3 and Questions 6 and 7 have a condition like "$x>1$": so that the value of secant is positive and $\tan\theta\ge0$.

Completing the square

If you see $\sqrt{ax^2+bx+c}$ or $\dfrac{1}{ax^2+bx+c}$, first complete the square, then reduce it to one of the three types above.

Example 4. $\displaystyle\int\frac{dx}{\sqrt{x^2+2x+5}}$.

$x^2+2x+5=(x+1)^2+4$. With $u=x+1$ it reached the form of Example 2:

$\int\frac{du}{\sqrt{u^2+4}}=\ln\left(u+\sqrt{u^2+4}\right)+C=\ln\left(x+1+\sqrt{x^2+2x+5}\right)+C$

If the leading coefficient is negative ($5+4x-x^2$), first take the minus out: $-(x^2-4x)+5=-(x-2)^2+4+5=9-(x-2)^2$. This is the sine type.

Algebraic inverse substitution

In Adams, “inverse substitution” is not only trigonometric. Writing $x$ as a power to get rid of a root is the same idea.

Example 5. $\displaystyle\int\frac{\sqrt x}{1+x}\,dx$. $x=u^2$, $dx=2u\,du$:

$\int\frac{u\cdot2u}{1+u^2}du=2\int\frac{u^2}{1+u^2}du=2\int\left(1-\frac{1}{1+u^2}\right)du=2u-2\arctan u+C$

$u=\sqrt x$: $2\sqrt x-2\arctan\sqrt x+C$. In the integrand that became rational, the numerator’s degree equals the denominator’s, so we divided first (week 13).

Which method? A decision table

is there a root in the integrand?
 ├─ the derivative of the radicand (up to a constant) is outside the root  -> u = radicand (simple)
 ├─ a^2 - x^2                                        -> x = a sin θ
 ├─ a^2 + x^2                                        -> x = a tan θ
 ├─ x^2 - a^2                                        -> x = a sec θ
 ├─ ax^2 + bx + c                                    -> complete the square, then the three above
 └─ √x, ∛x                                           -> x = u^2, u^3

PITFALL · an unnecessary trigonometric substitution. For $\displaystyle\int\frac{x\,dx}{\sqrt{x^2+9}}$, $x=3\tan\theta$ works but takes four lines. With $u=x^2+9$ it is $\sqrt{x^2+9}+C$ in one line. What saves time on an exam is seeing the simple way first.

Common mistakes