01 · The absolute-value identity
What is √(16cos²θ) for an unrestricted real θ?
Hint
A square root is nonnegative.
Worked solution
4|cos θ|, not automatically 4cos θ.
Understand · explore · practise
Trigonometric substitution and inverse trig integrals
Use x = a sin theta and x = a tan theta to integrate radicals and quadratic denominators. Derive arcsin and arctan integrals, choose valid branches and evaluate exact definite integrals.
Before you startIntegration by substitution, radians, trig identities and inverse trig functions.
01 / Choose an identity that simplifies the expression
Sine and tangent substitutions remove different quadratic forms.
a² − x²: try x = a sin θ
Then a² − x² = a²cos²θ. For a > 0, its square root is a|cos θ|.
a² + x²: try x = a tan θ
Then a² + x² = a²sec²θ and dx = a sec²θ dθ.
Always include the differential and specify a suitable range for θ. Inverse trig functions use principal branches; they do not undo a trig function for every possible angle.
See why the branch mattersExplore
x = 1.5; a cos θ = 2.598076; √(a² − x²) = 2.598076.
On −π/2 < θ < π/2, cos θ is positive, so √(a² − x²) = a cos θ.
arcsin(x/a) = π/6, recovering θ.
The picture rescales each radius to fit. The square root is always a nonnegative length; the signed horizontal coordinate can be negative.
02 / Choose the sine branch before cancelling
The square root of cos squared is absolute cosine.
Simplify ∫1/√(9 − x²) dx using x = 3sin θ.Worked example
Choose −π/2 < θ < π/2
This covers −3 < x < 3 and makes cos θ positive.
dx = 3cos θ dθ; √(9 − x²) = 3cos θ
The branch makes the square-root simplification valid.
∫[3cos θ/(3cos θ)] dθ = ∫1 dθ
The factors cancel on the open interval.
θ + C = arcsin(x/3) + C
Use the principal inverse sine.
Watch: a signed coordinate is not a square-root length
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Derive the arcsin integral
The scale a cancels when it appears in the radical.
∫1/√(a² − x²) dx = arcsin(x/a) + C
a > 0; −a < x < a.
Check the formula by differentiation.Worked example
d/dx arcsin(x/a) = (1/a)/√(1 − x²/a²)
Use the chain rule.
√(1 − x²/a²) = √(a² − x²)/a
This uses a > 0.
The derivative is 1/√(a² − x²)
There is no extra outside factor of 1/a in this particular formula.
02 · A standard radical
Find ∫1/√(25 − x²) dx.
Hint
Here a = 5.
Worked solution
arcsin(x/5) + C, for −5 < x < 5.
03 · A constant numerator
Find ∫4/√(4 − x²) dx.
Hint
Keep the numerator outside.
Worked solution
4arcsin(x/2) + C, for −2 < x < 2.
04 / Handle a coefficient on x
Use a substitution or check the chain factor explicitly.
Find ∫1/√(25 − 4x²) dx.Worked example
Let v = 2x, so dx = dv/2
First remove the coefficient on x².
(1/2)∫1/√(25 − v²) dv
Now use the standard arcsin form.
(1/2)arcsin(2x/5) + C
Valid for −5/2 < x < 5/2.
04 · A scaled radical
Find ∫3/√(9 − 16x²) dx.
Hint
Use v = 4x.
Worked solution
(3/4)arcsin(4x/3) + C, for |x| < 3/4.
05 · Check a wrong factor
A student gives ∫1/√(9 − x²) dx = (1/3)arcsin(x/3) + C. What is wrong?
Hint
Differentiate the proposed answer.
Worked solution
The extra factor 1/3 makes the derivative too small. The answer is arcsin(x/3) + C.
05 / Use tangent for a sum of squares
The secant-squared factors cancel.
Find ∫1/(9 + x²) dx using x = 3tan θ.Worked example
Choose −π/2 < θ < π/2
Tangent is one-to-one and covers every real x.
dx = 3sec²θ dθ; 9 + x² = 9sec²θ
Use 1 + tan²θ = sec²θ.
(1/3)∫1 dθ
The scale factor now remains.
(1/3)arctan(x/3) + C
Return to x through the principal inverse tangent.
06 · Tangent differential
If x = 5tan θ, what is dx?
Hint
Differentiate with respect to θ.
Worked solution
5sec²θ dθ.
07 · A reciprocal quadratic
Find ∫2/(4 + x²) dx.
Hint
The standard scale is a = 2.
Worked solution
arctan(x/2) + C.
06 / Remember the arctan scale
The sum-of-squares denominator has no square root.
∫1/(a² + x²) dx = (1/a)arctan(x/a) + C
a > 0; valid for every real x.
Find ∫1/(4 + 9x²) dx.Worked example
Let v = 3x, so dx = dv/3
This gives (1/3)∫1/(4 + v²) dv.
(1/6)arctan(v/2) + C
The standard integral contributes another factor 1/2.
(1/6)arctan(3x/2) + C
Check its derivative before finishing.
08 · Shift the quadratic
Find ∫1/[4 + (x − 1)²] dx.
Hint
Use v = x − 1, then a = 2.
Worked solution
(1/2)arctan((x − 1)/2) + C.
09 · Complete the square
Find ∫1/(x² + 4x + 13) dx.
Hint
The denominator is (x + 2)² + 9.
Worked solution
(1/3)arctan((x + 2)/3) + C.
07 / Change trig bounds on the chosen branch
Use exact principal angles.
Evaluate ∫₀^(3/2) 1/√(9 − x²) dx.Worked example
x = 3sin θ, with −π/2 < θ < π/2
The interval is safely inside the radical domain.
x = 0 → θ = 0; x = 3/2 → θ = π/6
Use the principal angle, not 5π/6.
∫₀^(π/6) 1 dθ = π/6
The answer is exact.
Evaluate ∫₀³ 1/(9 + x²) dx.Worked example
x = 3tan θ maps 0 to 0 and 3 to π/4
Stay on the principal tangent branch.
(1/3)∫₀^(π/4) dθ = π/12
Keep the scale factor.
10 · An exact inverse-tangent value
Evaluate ∫₀² 1/(4 + x²) dx.
Hint
Use (1/2)arctan(x/2).
Worked solution
π/8.
08 / Integrate the radical itself
Sine substitution can leave a cos-squared integral.
Find ∫√(9 − x²) dx for −3 < x < 3.Worked example
x = 3sin θ; dx = 3cos θ dθ
Use the principal sine branch.
∫9cos²θ dθ
The radical and differential each contribute a cosine.
(9/2)θ + (9/4)sin(2θ) + C
Use cos²θ = (1 + cos 2θ)/2.
(x/2)√(9 − x²) + (9/2)arcsin(x/3) + C
Replace sin 2θ with 2sin θ cos θ, then return to x.
11 · A general radical primitive
For a > 0, state a primitive of √(a² − x²) on |x| < a.
Hint
Follow the worked example with scale a.
Worked solution
(x/2)√(a² − x²) + (a²/2)arcsin(x/a) + C.
12 · A sine-squared remainder
Find ∫x²/√(9 − x²) dx.
Hint
x = 3sin θ gives 9∫sin²θ dθ.
Worked solution
(9/2)arcsin(x/3) − (x/2)√(9 − x²) + C, for |x| < 3.
09 / Do not undo sine outside its principal branch
The same x can correspond to different angles.
At θ = 2π/3, sin θ = √3/2 but cos θ is negative. With x = 3sin θ, the quantity √(9 − x²) is positive and equals −3cos θ at that angle. Also arcsin(sin(2π/3)) = π/3, not 2π/3.
Choosing −π/2 < θ < π/2 from the beginning makes the radical simplification and reverse substitution agree. At θ = ±π/2, the reciprocal-radical integrand is undefined; do not cancel zero factors there without separate endpoint analysis.
13 · Choose the correct inverse angle
What is arcsin(sin(2π/3))?
Hint
Inverse sine returns an angle in [−π/2, π/2].
Worked solution
π/3.
10 / Use the simplest suitable substitution
A numerator may make trig substitution unnecessary.
Find ∫x/√(9 − x²) dx.Worked example
Use u = 9 − x²; du = −2x dx
The numerator already supplies the inner derivative.
−(1/2)∫u⁻¹ᐟ² du
This is a direct power integral.
−√(9 − x²) + C
No inverse trig function is needed.
The denominator shape suggests a method, but inspect the whole integrand. A factor x often creates a simpler reverse-chain or logarithmic substitution.
14 · Choose between log and arctan
Find ∫2x/(9 + x²) dx.
Hint
The numerator is the denominator’s derivative.
Worked solution
ln(9 + x²) + C, using u = 9 + x².
15 · Name the appropriate identity
Which identity simplifies 16 + x² after x = 4tan θ?
Hint
Factor out 16.
Worked solution
1 + tan²θ = sec²θ, so 16 + x² = 16sec²θ.
11 / Match the quadratic form and check the branch
Carry the scale factor through the differential.
- Use sine for a² − x² and tangent for a² + x² when appropriate.
- State a range where the inverse and square-root simplifications are valid.
- Transform dx and every remaining factor.
- Use trig identities for any remaining squares.
- Return to x or transform definite bounds consistently.
- Differentiate the final answer to check scale and sign.
16 · Combine a shift and scale
Find ∫1/√(4 − (x − 1)²) dx.
Hint
First use v = x − 1, then the arcsin form with a = 2.
Worked solution
arcsin((x − 1)/2) + C, for −1 < x < 3.
Section 1 of 11 · Choose an identity that simplifies the expression