Tags: integration techniques, u-substitution, integration by parts, inverse trig integrals, trig integrals, power-reduction identity, trigonometric substitution, calculus practice, antiderivatives
This set covers six integrals that test your ability to identify and apply core integration techniques: inverse trig integration by parts, basic u-substitution, logarithmic integration by parts, trig power manipulation, trigonometric substitution, and power-reduction identities. The single most important skill here is recognising which technique to reach for before you start computing.
Integration by parts
A method based on the product rule: ∫ u dv = uv − ∫ v du. Use when the integrand is a product of two different "types" of function (e.g. polynomial × logarithm, polynomial × inverse trig).
u-substitution (change of variable)
Reversing the chain rule. Spot an inner function whose derivative also appears in the integrand, then substitute to simplify.
Power-reduction identity
A trig identity that lowers the power of sin or cos by rewriting in terms of a double angle. The key ones: cos²θ = (1 + cos 2θ)/2 and sin²θ = (1 − cos 2θ)/2.
Trigonometric substitution (trig sub)
A technique for integrands containing √(x² − a²), √(a² − x²), or √(x² + a²). You substitute x = a sec θ, x = a sin θ, or x = a tan θ respectively, converting the root into a trig expression.
LIATE rule
A mnemonic for choosing u in integration by parts. Prioritise (from first to last): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. The function earlier in the list becomes u.
Technique: Integration by parts. The integrand is a single inverse trig function with no obvious substitution, so you pair it with dx.
Let u = cos⁻¹(x), so du = −1/√(1 − x²) dx
Let dv = dx, so v = x
Applying the formula:
∫ cos⁻¹(x) dx = x cos⁻¹(x) − ∫ x · (−1/√(1 − x²)) dx
= x cos⁻¹(x) + ∫ x / √(1 − x²) dx
For the remaining integral, use u-sub: let w = 1 − x², dw = −2x dx.
∫ x / √(1 − x²) dx = −(1/2) ∫ w^(−1/2) dw = −√(1 − x²)
Final answer:
∫ cos⁻¹(x) dx = x cos⁻¹(x) − √(1 − x²) + C
Technique: Straightforward u-substitution. When you see sinⁿx · cos x (or cosⁿx · sin x), the cosine (or sine) is the derivative of the other factor.
Let u = sin x, du = cos x dx
∫ sin³x cos x dx = ∫ u³ du = u⁴/4
Final answer:
∫ sin³x cos x dx = sin⁴x / 4 + C
Technique: Integration by parts. The LIATE rule tells you to let u = ln x (logarithmic beats algebraic).
Let u = ln x, so du = 1/x dx
Let dv = x dx, so v = x²/2
∫ x ln x dx = (x²/2) ln x − ∫ (x²/2)(1/x) dx
= (x²/2) ln x − (1/2) ∫ x dx
= (x²/2) ln x − x²/4
Final answer:
∫ x ln x dx = (x²/2) ln x − x²/4 + C
Technique: Rewrite the trig powers so you can pull out a sec x tan x factor (the derivative of sec x), then substitute.
Start by rewriting tan³x = tan²x · tan x = (sec²x − 1) tan x:
∫ sec³x tan³x dx = ∫ sec³x (sec²x − 1) tan x dx
= ∫ sec²x (sec²x − 1) · sec x tan x dx
Now let u = sec x, du = sec x tan x dx:
= ∫ u² (u² − 1) du = ∫ (u⁴ − u²) du
= u⁵/5 − u³/3
Final answer:
∫ sec³x tan³x dx = sec⁵x / 5 − sec³x / 3 + C
Technique: Trig substitution. The form √(x² − a²) with a = 3 calls for x = 3 sec θ.
Let x = 3 sec θ, so dx = 3 sec θ tan θ dθ
√(x² − 9) = √(9 sec²θ − 9) = 3 tan θ
x² = 9 sec²θ
Substituting:
∫ 1 / (9 sec²θ · 3 tan θ) · 3 sec θ tan θ dθ
= ∫ 1 / (9 sec θ) dθ = (1/9) ∫ cos θ dθ
= (1/9) sin θ
To convert back: from x = 3 sec θ, we get cos θ = 3/x, and sin θ = √(x² − 9)/x (draw the reference triangle).
Final answer:
∫ 1 / (x² √(x² − 9)) dx = √(x² − 9) / (9x) + C
Technique: Repeated use of the power-reduction identity. There is no convenient u-sub when you have an even power of cosine with nothing else around.
Step 1: Write cos⁴(2x) = [cos²(2x)]².
Apply cos²(2x) = (1 + cos 4x)/2:
cos⁴(2x) = [(1 + cos 4x)/2]² = (1 + 2 cos 4x + cos²4x) / 4
Step 2: Reduce cos²(4x) = (1 + cos 8x)/2:
= (1 + 2 cos 4x + (1 + cos 8x)/2) / 4
= (2 + 4 cos 4x + 1 + cos 8x) / 8
= (3 + 4 cos 4x + cos 8x) / 8
Step 3: Integrate term by term:
∫ cos⁴(2x) dx = (1/8) [3x + sin 4x + sin 8x / 8]
Final answer:
∫ cos⁴(2x) dx = 3x/8 + sin(4x)/8 + sin(8x)/64 + C
Integration by parts: ∫ u dv = uv − ∫ v du
Power-reduction identities:
cos²θ = (1 + cos 2θ) / 2
sin²θ = (1 − cos 2θ) / 2
Trig substitution reference:
√(x² − a²) → x = a sec θ
√(a² − x²) → x = a sin θ
√(x² + a²) → x = a tan θ
Pythagorean identity for sec/tan: tan²x = sec²x − 1
⚠️ For problem (a), the trick of treating a lone inverse trig function as u with dv = dx is a classic exam pattern. If you see a single arccos, arcsin, or arctan with nothing else, integration by parts is the move.
⚠️ Problem (b) is a one-step u-sub. On timed exams this should take under a minute. If you are spending longer, you have likely over-complicated it.
⚠️ In problem (d), the key insight is factoring out sec x tan x as a unit. Any time you have odd powers of both sec and tan, try to reserve one sec x tan x for du.
⚠️ Problem (f) is the most algebraically tedious type. Double-check your double-angle arguments: cos²(2x) reduces to cos(4x), then cos²(4x) reduces to cos(8x). Losing track of the argument is the most common error.
⚠️ For trig sub problems (e), always draw the reference triangle to convert back to x. Forgetting to back-substitute is a frequent mark-losing mistake.
Q: You see ∫ arctan(x) dx on an exam. What is your first move?
A: Integration by parts with u = arctan(x), dv = dx. This is the same pattern as problem (a): a lone inverse trig function.
Q: When you have ∫ sinⁿx cos x dx with n a positive integer, what substitution immediately applies?
A: Let u = sin x, du = cos x dx. The integral becomes ∫ uⁿ du = uⁿ⁺¹/(n+1) + C.
Q: What identity do you use to integrate an even power of cosine, like cos⁴θ?
A: The power-reduction identity cos²θ = (1 + cos 2θ)/2, applied repeatedly. Square the expression and reduce again until no powers remain.
Q: In problem (e), why choose x = 3 sec θ rather than x = 3 sin θ or x = 3 tan θ?
A: The radical is √(x² − 9), which matches the form √(x² − a²). That form calls for x = a sec θ so the square root simplifies via sec²θ − 1 = tan²θ.
integration by parts, LIATE rule, u-substitution, change of variable, inverse trig integral, arccos integral, trig powers, odd trig power, even trig power, power-reduction formula, double angle identity, half angle identity, trigonometric substitution, sec substitution, reference triangle, back-substitution, calculus 2, MATH 151, MATH 152, Texas A&M, antiderivative techniques