Guides / Integrals

Trig substitution: pick sin, tan, or sec from the square root, then draw the triangle

Trig substitution handles integrals with √(a² − x²), √(a² + x²) or √(x² − a²) in them, where an ordinary u-substitution has nothing to grab. The right substitution turns the square root into a single trig function, and a right triangle turns the answer back into x. The mistakes are in the bookkeeping: the dx, the limits, and the triangle.

Why it works

Each substitution is built so that a Pythagorean identity swallows the square root. With x = 3 sin θ, 9 − x² = 9 − 9 sin²θ = 9 cos²θ, and the square root of that is 3 cos θ. The other two use 1 + tan²θ = sec²θ and sec²θ − 1 = tan²θ the same way.

To get back to x, read the substitution as a triangle. x = 3 sin θ says sin θ = x/3: opposite side x, hypotenuse 3. Pythagoras gives the adjacent side, √(9 − x²), and then every trig function of θ can be read off the triangle. The restricted ranges of θ (for sin, from −π/2 to π/2) keep cos θ positive, so the square root really is 3 cos θ and not −3 cos θ.

Example 1: a² − x², x = a sin θ

Find ∫ dxx²√(9 − x²).

1. x = 3 sin θ, dx = 3 cos θ dθ, √(9 − x²) = 3 cos θa² − x² with a = 3, so use sin.
2. ∫ 3 cos θ dθ9 sin²θ · 3 cos θ = 19∫ csc²θ dθThe cos θ from dx cancels the one from the root.
3. = −19 cot θ + CThe derivative of cot θ is −csc²θ.
4. = −√(9 − x²)9x + CThe triangle has opposite x, hypotenuse 3 and adjacent √(9 − x²), and cot θ is adjacent over opposite.

Example 2: a² + x², x = a tan θ

Find ∫ dx(x² + 4)^3/2.

1. x = 2 tan θ, dx = 2 sec²θ dθ, (x² + 4)^3/2 = (2 sec θ)³ = 8 sec³θ4 tan²θ + 4 = 4 sec²θ, and its square root is 2 sec θ.
2. ∫ 2 sec²θ8 sec³θ dθ = 14∫ cos θ dθ = 14 sin θ + C1/sec θ = cos θ.
3. = x4√(x² + 4) + Ctan θ = x/2: opposite x, adjacent 2, hypotenuse √(x² + 4). sin θ is opposite over hypotenuse.

Example 3: x² − a², x = a sec θ

Find ∫ √(x² − 9)x dx for x ≥ 3.

1. x = 3 sec θ, dx = 3 sec θ tan θ dθ, √(x² − 9) = 3 tan θx² − a² with a = 3, so use sec. For x ≥ 3, θ is between 0 and π/2, where tan θ ≥ 0.
2. ∫ 3 tan θ3 sec θ · 3 sec θ tan θ dθ = 3∫ tan²θ dθThe sec θ cancels.
3. = 3∫ (sec²θ − 1) dθ = 3 tan θ − 3θ + Ctan²θ has no direct antiderivative, but sec²θ − 1 does.
4. = √(x² − 9) − 3 arcsec(x/3) + C3 tan θ is √(x² − 9) from step 1, and θ = arcsec(x/3) from the substitution.

Example 4: a definite integral, limits in θ

Evaluate ∫₀² √(4 − x²) dx.

1. x = 2 sin θ: x = 0 → θ = 0, x = 2 → θ = π2Change the limits along with the variable, and there's no need to go back to x.
2. ∫₀^π/2 2 cos θ · 2 cos θ dθ = ∫₀^π/2 4 cos²θ dθThe root is 2 cos θ, and dx = 2 cos θ dθ.
3. = ∫₀^π/2 2(1 + cos 2θ) dθ = [2θ + sin 2θ]₀^π/2The double-angle identity cos²θ = (1 + cos 2θ)/2.
4. = πA check: y = √(4 − x²) is a quarter of a circle of radius 2, and its area is ¼ · π · 2² = π.

Common mistakes, and the exact line they happen on

1. Substituting for x but not for dx

The quarter circle from Example 4.

1. √(4 − x²) = 2 cos θ, limits 0 to π2Right so far.
2. ∫₀^π/2 2 cos θ dθ = 2This is the mistake. dx became dθ, but dx = 2 cos θ dθ. Without that factor the answer is 2, not the area of the quarter circle, π ≈ 3.142.

2. Keeping the x-limits after changing to θ

Example 4 again.

1. ∫ 4 cos²θ dθ = 2θ + sin 2θRight so far.
2. [2θ + sin 2θ]₀² = 4 + sin 4 ≈ 3.243This is the mistake. 0 and 2 are values of x. In θ the limits are 0 and π/2, which give π ≈ 3.142. Either change the limits, or go back to x before you evaluate. Don't mix the two.

3. Reading the wrong side off the triangle

The last step of Example 1.

1. −19 cot θ + C, with sin θ = x3Right so far.
2. cot θ = x√(9 − x²), so the answer is −x9√(9 − x²) + CThis is the mistake. That's tan θ, opposite over adjacent. cot θ is adjacent over opposite, √(9 − x²)/x. Differentiating catches it: at x = 1 the integrand is 1/√8 ≈ 0.354, but the derivative of this answer is about −0.044. Label the triangle's sides before you read anything off it.

Practice

1. Find ∫ dx√(16 − x²).

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   x = 4 sin θ gives ∫ dθ = θ + C = arcsin(x/4) + C.

2. Find ∫ dxx² + 9.

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   x = 3 tan θ gives 13∫ dθ = 13 arctan(x/3) + C. No square root, but the same substitution works.

3. Find ∫ x²√(1 − x²) dx.

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   x = sin θ gives ∫ sin²θ dθ = θ2 − sin θ cos θ2, which is ½(arcsin x − x√(1 − x²)) + C.

4. Evaluate ∫₀¹ dx(x² + 1)^3/2.

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   x = tan θ, θ from 0 to π/4: ∫ cos θ dθ = sin(π/4) = √22 ≈ 0.707.

5. Find ∫ dx√(x² − 1) for x > 1.

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   x = sec θ gives ∫ sec θ dθ = ln |sec θ + tan θ|, which is ln(x + √(x² − 1)) + C.