Tags: trig integrals, algebraic simplification, definite integrals, trig substitution, trig identity, tan and sec integrals, calculus practice, antiderivatives
Part 2 covers five more integrals emphasising trig power manipulation with tan/sec, algebraic simplification before integrating, definite integrals requiring trig identities, and two flavours of trig substitution (√(x² − a²) and √(x² + a²)). Several of these reward a moment of rewriting before diving in.
Algebraic simplification (before integrating)
Sometimes the fastest technique is no technique at all. Dividing through, splitting fractions, or cancelling before you integrate can turn a scary-looking integrand into basic power-rule terms.
Pythagorean identity (sin²x + cos²x = 1)
Used to rewrite sin²x as 1 − cos²x (or vice versa) so you can split an integral into simpler pieces. Essential whenever you need to convert between sin and cos powers.
Definite integral
An integral with specific bounds. Compute the antiderivative, then evaluate at the upper and lower limits. For trig-sub problems, you can either convert the limits to θ-limits or back-substitute into x before evaluating.
Reference triangle
A right triangle drawn from the trig substitution relationship (e.g. x = a sec θ implies adjacent = a, hypotenuse = x, opposite = √(x² − a²)). Used to convert the θ-expression back to x.
Technique: Rewrite powers and reserve sec x tan x for the substitution. When you have an odd power of tan alongside sec, pull out one sec x tan x factor.
Rewrite: tan³x sec⁶x = tan²x · sec⁵x · (sec x tan x)
Since tan²x = sec²x − 1:
= (sec²x − 1) sec⁵x · sec x tan x dx
Let u = sec x, du = sec x tan x dx:
= ∫ (u² − 1) u⁵ du = ∫ (u⁷ − u⁵) du
= u⁸/8 − u⁶/6
Final answer:
∫ tan³x sec⁶x dx = sec⁸x / 8 − sec⁶x / 6 + C
Technique: No special method needed. Divide each term in the numerator by 4x, then integrate term by term.
(3x² + 2x + 1) / (4x) = 3x/4 + 1/2 + 1/(4x)
Now integrate:
∫ [3x/4 + 1/2 + 1/(4x)] dx = 3x²/8 + x/2 + (1/4) ln|x|
Final answer:
∫ (3x² + 2x + 1) / (4x) dx = 3x²/8 + x/2 + (1/4) ln|x| + C
This is a good reminder to always check whether simple algebra eliminates the need for a technique.
Technique: Use sin²x = 1 − cos²x to rewrite sin³x, then split the fraction. The hint on the problem itself says "may need to use an identity to simplify."
sin³x / cos x = sin x · sin²x / cos x = sin x (1 − cos²x) / cos x
= sin x / cos x − sin x cos x
= tan x sin x − sin x cos x
Wait, let's be a bit more careful. Splitting directly:
= sin x / cos x − sin x cos²x / cos x
= sin x / cos x − sin x cos x
Actually cleaner: write it as (sin x)(1 − cos²x) / cos x = sin x / cos x − sin x cos x = tan x sin x − sin x cos x. Hmm, let's take the simpler path.
Rewrite:
∫ sin³x / cos x dx = ∫ sin x (1 − cos²x) / cos x dx
= ∫ sin x / cos x dx − ∫ sin x cos x dx
= ∫ tan x sin x dx ... this is getting circular. Better approach:
Let u = cos x, du = −sin x dx:
∫ sin x (1 − cos²x) / cos x dx = −∫ (1 − u²) / u du
= −∫ (1/u − u) du = −[ln|u| − u²/2]
= −ln|cos x| + cos²x / 2
Evaluate from 0 to π/4:
At x = π/4: −ln(cos π/4) + cos²(π/4)/2 = −ln(√2/2) + (1/2)/2 = −ln(√2/2) + 1/4
At x = 0: −ln(1) + 1/2 = 0 + 1/2
Result: [−ln(√2/2) + 1/4] − [1/2]
= −ln(√2/2) − 1/4
Since −ln(√2/2) = ln(2/√2) = ln(√2) = (1/2) ln 2:
Final answer:
∫₀^(π/4) sin³x / cos x dx = (1/2) ln 2 − 1/4
Technique: Trig substitution. The form √(x² − 1) with a = 1 calls for x = sec θ.
Let x = sec θ, dx = sec θ tan θ dθ
√(x² − 1) = tan θ
x⁴ = sec⁴θ
∫ tan θ / sec⁴θ · sec θ tan θ dθ = ∫ tan²θ / sec³θ dθ
= ∫ (sin²θ / cos²θ) · cos³θ dθ = ∫ sin²θ cos θ dθ
Let w = sin θ, dw = cos θ dθ:
= ∫ w² dw = w³/3 = sin³θ / 3
Back-substitute using the reference triangle (hypotenuse = x, adjacent = 1, opposite = √(x² − 1)):
sin θ = √(x² − 1) / x
Final answer:
∫ √(x² − 1) / x⁴ dx = (x² − 1)^(3/2) / (3x³) + C
Technique: Trig substitution. The form √(1 + x²) calls for x = tan θ.
Let x = tan θ, dx = sec²θ dθ
√(1 + x²) = √(1 + tan²θ) = sec θ
The integral becomes:
∫ sec θ / tan θ · sec²θ dθ = ∫ sec³θ / tan θ dθ
Rewrite in sin and cos:
= ∫ (1/cos³θ) · (cos θ / sin θ) dθ = ∫ 1 / (cos²θ sin θ) dθ
= ∫ sec²θ / sin θ dθ = ∫ (1 + tan²θ) / sin θ dθ ... this path is getting tangled.
Alternative rewrite using sec²θ = 1 + tan²θ and csc θ = 1/sin θ:
∫ 1 / (cos²θ sin θ) dθ
Use the identity 1/cos²θ = sec²θ and write sin θ in the denominator:
= ∫ sec²θ · csc θ dθ
Split using 1 = sin²θ + cos²θ in the numerator of 1/(cos²θ sin θ):
= ∫ sin θ / cos²θ dθ + ∫ 1/sin θ dθ ... no, let's be more careful.
Actually: 1/(cos²θ sin θ) = sin θ/(cos²θ sin²θ)... let's use partial fractions in trig. The cleanest method:
Write 1/(cos²θ sin θ) = (sin²θ + cos²θ)/(cos²θ sin θ) = sin θ/cos²θ + 1/sin θ
So: = ∫ sin θ / cos²θ dθ + ∫ csc θ dθ
First integral: let w = cos θ, dw = −sin θ dθ → −∫ w⁻² dw = 1/cos θ = sec θ
Second integral: ∫ csc θ dθ = −ln|csc θ + cot θ|
Combining: sec θ − ln|csc θ + cot θ| + C
Back-substitute from the reference triangle (opposite = x, adjacent = 1, hypotenuse = √(1 + x²)):
sec θ = √(1 + x²)
csc θ = √(1 + x²)/x
cot θ = 1/x
Final answer:
∫ √(1 + x²) / x dx = √(1 + x²) − ln| (√(1 + x²) + 1) / x | + C
This can also be written as:
= √(1 + x²) + ln|x| − ln(1 + √(1 + x²)) + C
Pythagorean identities:
sin²x + cos²x = 1
1 + tan²x = sec²x
1 + cot²x = csc²x
Standard results used in these problems:
∫ tan x dx = −ln|cos x| + C = ln|sec x| + C
∫ csc θ dθ = −ln|csc θ + cot θ| + C
d/dx [sec x] = sec x tan x
Trig substitution reference triangles:
x = sec θ → opposite = √(x² − 1), adjacent = 1, hypotenuse = x
x = tan θ → opposite = x, adjacent = 1, hypotenuse = √(1 + x²)
⚠️ Problem (h) is the sort of question that separates students who think before computing from those who don't. Always check whether you can simplify algebraically before reaching for a technique. Splitting the fraction saves several minutes compared to attempting partial fractions or other methods.
⚠️ For problems like (g) with mixed tan and sec powers: if the power of tan is odd, save one sec x tan x for du and convert remaining tan² to sec² − 1. If the power of sec is even, save sec²x for du and convert remaining sec² to 1 + tan².
⚠️ Problem (i) is a definite integral. You can either change the limits when you substitute (x = 0 → u = cos 0 = 1; x = π/4 → u = cos π/4 = √2/2), or find the antiderivative in x and evaluate. Changing limits avoids back-substitution entirely.
⚠️ In problems (j) and (k), the back-substitution step via the reference triangle is where most errors happen. Draw the triangle, label all three sides, and read off the trig function you need.
⚠️ Problem (k) is one of the hardest in this set. The intermediate algebra after trig sub is messy, and the splitting trick (writing 1 = sin²θ + cos²θ in the numerator) is not obvious. If you see this type on an exam, partial credit for setting up the substitution correctly is usually available even if the algebra goes sideways.
Q: You see ∫ (5x³ − x + 2) / (3x²) dx. What should you do first?
A: Divide every term by 3x²: (5x/3) − 1/(3x) + 2/(3x²). Then integrate term by term with the power rule and ln. No advanced technique needed.
Q: You have ∫ tan⁵x sec⁴x dx. Which factor do you reserve for du, and what substitution do you make?
A: The power of sec is even, so save sec²x for du and let u = tan x, du = sec²x dx. Rewrite the remaining sec² as 1 + tan² and you have a polynomial in u.
Q: For ∫ √(x² − 4) / x³ dx, what substitution should you use and what does each piece become?
A: x = 2 sec θ, dx = 2 sec θ tan θ dθ, √(x² − 4) = 2 tan θ, x³ = 8 sec³θ.
Q: How do you evaluate a definite integral after trig substitution without back-substituting?
A: Convert the x-limits to θ-limits at the start. For example, if x = a sec θ and x = a at the lower bound, then sec θ = 1, so θ = 0. Evaluate the θ-antiderivative directly at the θ-limits.
tan and sec integrals, trig power integrals, odd power of tangent, even power of secant, algebraic simplification, splitting fractions, definite integral evaluation, trig substitution, sec substitution, tan substitution, reference triangle, back-substitution, Pythagorean identity, sin squared plus cos squared, csc integral, calculus 2, MATH 151, MATH 152, Texas A&M, integration practice