Trig sub

Draw the triangle

Trig sub means you build a right triangle out of the pieces already in the integrand. One acute angle is θ. You read sin, cos, or tan off the sides, differentiate to get dθ, integrate in θ, then translate every trig function back into x on that same triangle.

Thursday only needs two pictures. If the algebra is a² + u², the adjacent side is a and the opposite side is u. That angle is an arctangent. If the algebra is a² − u² under a square root, the opposite side is u and the hypotenuse is a. That angle is an arcsine.

Arctan keeps a factor of 1/a. Arcsin does not. In the arcsine substitution the a on du and the a on the root cancel. In the arctangent substitution one a is left in the denominator.

When the integral is exactly du/(a² + u²) or du/√(a² − u²), the triangle collapses to one line. An extra power of x, or a bare √(a² − x²) with nothing under it, has to be substituted all the way through θ. Draw the triangle on the quiz even when you already know the one-line answer. The derivation is the work.

Which triangle

Match the algebra first. Then decide whether you are allowed to stop.

du / (a² + u²)

(1/a) arctan(u/a) + C

du / √(a² − u²)

arcsin(u/a) + C

√(a² − u²) du

sin θ, then cos²

u² / √(a² − u²) du

sin θ, then sin²

du / (u² √(a² + u²))

tan θ, then a u-sub in θ

A lone √(x + 4) is not one of these. Let u = x + 4 and rewrite the leftover x as u minus a constant. No triangle.

√(u² − a²) is the third textbook triangle, hypotenuse u and adjacent side a, so u = a sec θ. Nothing on the Sep 24 practice uses it. Skip it Thursday.

a² + u²

Adjacent side a along the bottom. Opposite side u up the vertical. Hypotenuse √(a² + u²). θ sits next to side a.

tan θ = u/a sec θ = √(a² + u²) / a

The reference case is a = 1 and u = x, which is ∫ dx/(1 + x²).

tan θ = x sec θ = √(1 + x²)

Differentiate tan θ = x.

sec² θ · dθ/dx = 1 dθ/dx = 1/sec² θ = 1/(1 + tan² θ) = 1/(1 + x²)

So the integral is the angle.

∫ dx/(1 + x²) = θ + C = arctan x + C

Why the 1/a shows up

Same triangle, general u = a tan θ. Then du = a sec² θ dθ, and a² + u² = a² sec² θ.

∫ a sec² θ dθ / (a² sec² θ) = ∫ dθ / a = θ/a ∫ du/(a² + u²) = (1/a) arctan(u/a) + C

For ∫ dx/(x² + 16), the adjacent side is 4 and the opposite side is x. tan θ = x/4.

sec² θ · dθ/dx = 1/4 dθ/dx = (1/4) / (1 + (x/4)²) = 4/(x² + 16)

The 4 in the numerator is what the derivative produced, so you divide it back out.

∫ dx/(x² + 16) = (1/4) arctan(x/4) + C

When arctan is not enough

∫ dx / (x² √(1 + x²)) uses the same triangle as the arctan derivation. Opposite x, adjacent 1, hypotenuse √(1 + x²). You cannot stop at dθ/dx. The x² in the denominator does not cancel into 1 + x².

x = tan θ dx = sec² θ dθ √(1 + x²) = sec θ ∫ sec² θ dθ / (tan² θ · sec θ) = ∫ sec θ / tan² θ dθ

Rewrite sec and tan in sin and cos. The cosines cancel once.

sec θ / tan² θ = (1/cos θ) / (sin² θ / cos² θ) = cos θ / sin² θ ∫ cos θ / sin² θ dθ

Now it is an ordinary substitution. Let u = sin θ, so du = cos θ dθ.

∫ u⁻² du = −1/u + C = −1/sin θ + C

On the triangle, sin θ = opposite / hypotenuse = x / √(1 + x²). Flip it.

1/sin θ = √(1 + x²) / x ∫ dx / (x² √(1 + x²)) = −√(1 + x²) / x + C

a² − u²

Opposite side u. Hypotenuse a. Adjacent side √(a² − u²). The expression under the root has to stay positive, so u stays between −a and a.

sin θ = u/a cos θ = √(a² − u²) / a

For the plain form, substitute u = a sin θ. Then du = a cos θ dθ and √(a² − u²) = a cos θ.

∫ a cos θ dθ / (a cos θ) = ∫ dθ = θ ∫ du / √(a² − u²) = arcsin(u/a) + C

The a cancelled. That is why this answer has no 1/a in front.

∫ dx/√(25 − x²) is that formula with a = 5. You can also derive it the long way, which is what the quiz wants to see.

sin θ = x/5 cos θ · dθ/dx = 1/5 dθ/dx = (1/5) / cos θ = 1/√(25 − x²) ∫ dx/√(25 − x²) = arcsin(x/5) + C

You only get to stop there when the integrand is exactly that shape. √(4 − 9x²) still counts, after you rename 3x as u. A bare √(4 − x²), or x² on top of the root, does not.

√(4 − x²)

Opposite x, hypotenuse 2, adjacent √(4 − x²). The root is the whole integrand, so the cosines multiply instead of cancelling.

x = 2 sin θ dx = 2 cos θ dθ √(4 − x²) = 2 cos θ ∫ (2 cos θ)(2 cos θ) dθ = 4 ∫ cos² θ dθ

Start from the double-angle identity, the same one as the cos²(2x) problems. cos 2α = 2 cos² α − 1, so cos² α = (1 + cos 2α)/2.

4 ∫ (1 + cos 2θ)/2 dθ = 2θ + sin 2θ + C

sin 2θ = 2 sin θ cos θ, so the antiderivative is 2θ + 2 sin θ cos θ + C. Read both factors off the triangle.

sin θ = x/2 cos θ = √(4 − x²) / 2 2 sin θ cos θ = 2 · (x/2) · √(4 − x²)/2 = (x/2) √(4 − x²) ∫ √(4 − x²) dx = 2 arcsin(x/2) + (x/2) √(4 − x²) + C

Differentiate it if you have a minute. The product rule on the second term cancels the extra piece from the arcsin, and you land back on √(4 − x²).

x² on top of the root

∫ x² / √(9 − x²) dx. Opposite x, hypotenuse 3, adjacent √(9 − x²). This time a cosine cancels, and a sin² is left behind.

x = 3 sin θ dx = 3 cos θ dθ √(9 − x²) = 3 cos θ ∫ (9 sin² θ · 3 cos θ) / (3 cos θ) dθ = 9 ∫ sin² θ dθ

cos 2α = 1 − 2 sin² α, so sin² α = (1 − cos 2α)/2.

9 ∫ (1 − cos 2θ)/2 dθ = (9/2) θ − (9/4) sin 2θ + C

(9/4) sin 2θ = (9/4) · 2 sin θ cos θ = (9/2) sin θ cos θ. Put the sides in.

sin θ = x/3 cos θ = √(9 − x²) / 3 (9/2) sin θ cos θ = (9/2) · (x/3) · √(9 − x²)/3 = (x/2) √(9 − x²) ∫ x² / √(9 − x²) dx = (9/2) arcsin(x/3) − (x/2) √(9 − x²) + C

The sign in front of the triangle term is the whole difference between this integral and √(a² − x²). cos² produced a plus. sin² produces a minus.

Rescale before you draw

Pull constants inside the squared term so the triangle has clean sides. Then the formulas above apply to u, and you write u back as a multiple of x at the end.

4 − 9x²

4 − 9x² = 4 − (3x)². Let u = 3x, so du = 3 dx.

∫ dx / √(4 − 9x²) = (1/3) ∫ du / √(4 − u²) = (1/3) arcsin(u/2) + C = (1/3) arcsin(3x/2) + C

1 + 4x²

1 + 4x² = 1 + (2x)². For ∫ 4 dx / (1 + 4x²), let u = 2x, so du = 2 dx and dx = du/2.

∫ 4 · (du/2) / (1 + u²) = 2 arctan u + C = 2 arctan(2x) + C

A linear term in the quadratic

Complete the square when the quadratic does not factor into different linears. x² + 6x + 13 = (x + 3)² + 4. Let u = x + 3.

∫ dx / (x² + 6x + 13) = ∫ du / (u² + 4) = (1/2) arctan(u/2) + C = (1/2) arctan((x + 3)/2) + C

If the numerator is the derivative of that quadratic, stop. That integral is a log, and the triangle never starts. 2x + 5 over x² + 2x + 5 is both at once. Split the numerator as (2x + 2) + 3.

∫ (2x + 2)/(x² + 2x + 5) dx = ln(x² + 2x + 5)

The leftover 3 sits on (x + 1)² + 4.

3 ∫ du / (u² + 4) = (3/2) arctan((x + 1)/2) ln(x² + 2x + 5) + (3/2) arctan((x + 1)/2) + C

When the numerator is in the way

∫ (x + 1) / √(4 − x²) dx. Split it before you pick a method. The x is a substitution. The 1 is the triangle.

∫ x / √(4 − x²) dx + ∫ 1/√(4 − x²) dx

First piece. Let u = 4 − x², so du = −2x dx and x dx = −du/2.

−(1/2) ∫ u^(−1/2) du = −√(4 − x²)

Second piece is the arcsine form with a = 2.

arcsin(x/2) ∫ (x + 1) / √(4 − x²) dx = −√(4 − x²) + arcsin(x/2) + C

Do not feed the whole fraction into sin θ. The x on top wants to be the derivative of the inside.

Leave it alone

These are on the same practice sets. A triangle on any of them wastes the 45 minutes.

The numerator is a constant times the derivative of the inside. That is a log. (8x − 6)/(2x² − 3x + 5) is 2 ln|2x² − 3x + 5| + C.
Two different linear factors in the denominator. Partial fractions. Cover-up at each root.
√(x + 3) or √(x + 4) with an x in the numerator. Named substitution. Replace the leftover x. No triangle.
cos²(2x) or sin²(2x). Double-angle identity, then integrate. The triangle is for algebra in x, not for a squared trig function of x.
The derivative is off by a constant, like 2x times (x² + 2x). Write 2x as (2x + 2) − 2, substitute the matching piece, and integrate the leftover polynomial.

Every indefinite answer needs + C. Box the line in x, not the line still written in θ.