Indefinite Integration of Trigonometric, Exponential and Logarithmic Functions

Definition

• An antiderivative (inverse derivative) of f(x) on an interval is a function F(x) whose derivative is f(x), that is F′(x) = f(x) for every x in the interval.
• Adding any constant to an antiderivative gives another antiderivative, so every function has infinitely many antiderivatives.
• The set of all antiderivatives is called the indefinite integral and is written ∫ f(x) dx: ∫ f(x) dx = F(x) + C.
• Geometrically the indefinite integral is a family of parallel curves differing only by the constant, so at the same value of x their tangents all have the same slope.
• C is the constant of integration (arbitrary constant); it is fixed when a point on the curve is known.

Properties

• Integral of a sum or difference: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
• Constant factor: ∫ k f(x) dx = k ∫ f(x) dx, where k is a fixed real number.
• Power rule: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, n ≠ −1. The excluded case is exactly the logarithmic one: ∫ dx/x = ln|x| + C.
• Check any result by differentiating: d/dx of the answer must return the integrand.

Worked examples

1. ∫ (3x² + 4x + 5) dx = x³ + 2x² + 5x + C.
2. ∫ (x − 1)(x + 3) dx → expand first: ∫ (x² + 2x − 3) dx = x³/3 + x² − 3x + C.
3. ∫ (√x + 1/√x) dx = ∫ (x^1/2 + x^−1/2) dx = (2/3)x^3/2 + 2x^1/2 + C.
4. ∫ x(x² − 3)² dx: substitution is the efficient route (section 07) — u = x² − 3, du = 2x dx → ½∫ u² du = (x² − 3)³/6 + C. Expanding first also works: ∫ (x⁵ − 6x³ + 9x) dx = x⁶/6 − 3x⁴/2 + 9x²/2 + C, which differs from the factored form only by a constant.

Linear arguments

• Whenever the angle is (ax + b), divide by a: ∫ cos(ax + b) dx = sin(ax + b)/a + C, ∫ sin(ax + b) dx = −cos(ax + b)/a + C.
• ∫ sec²(2x) dx = tan(2x)/2 + C and ∫ e^3x dx = e^3x/3 + C — the same rule.
• Squares of trig functions are reduced with the double-angle identities first: sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2.
• Example: ∫ sin²x dx = ∫ (1 − cos 2x)/2 dx = x/2 − sin 2x/4 + C.

The two rules

• Natural base: ∫ e^u du = e^u + C, so for a linear exponent ∫ e^ax+b dx = e^ax+b/a + C.
• Any positive base a ≠ 1: ∫ a^u du = a^u/ln a + C, so ∫ a^kx dx = a^kx/(k ln a) + C.
• Reason: d/dx(a^x) = a^x ln a, therefore the factor ln a must be divided out when integrating.
• If the exponent is a function g(x), the chain rule must be undone: ∫ e^g(x) g′(x) dx = e^g(x) + C.

Worked examples

1. ∫ e^{2x − 1} dx = e^{2x − 1}/2 + C.
2. ∫ 3^x dx = 3^x/ln 3 + C.
3. ∫ x e^x² dx: let u = x², du = 2x dx → ½∫ e^u du = e^x²/2 + C.
4. ∫ (e^x + 2^x) dx = e^x + 2^x/ln 2 + C — integrate term by term, different rules for each.

The rule

• ∫ dx/x = ln|x| + C. The modulus is compulsory: for x < 0, d/dx ln(−x) = (−1)/(−x) = 1/x, so ln|x| is an antiderivative on both sides of zero.
• Linear argument: ∫ dx/(ax + b) = ln|ax + b|/a + C.
• Numerator equal to the derivative of the denominator: ∫ f′(x)/f(x) dx = ln|f(x)| + C.
• Multiples and powers: ∫ k·dx/x = k ln|x| + C = ln|x|^k + C, and ∫ dx/x² = −1/x + C (a power, not a logarithm).

Worked examples

1. ∫ dx/(3x − 5) = ln|3x − 5|/3 + C.
2. ∫ (2x + 1)/(x² + x + 4) dx: numerator is the derivative of the denominator → ln(x² + x + 4) + C (the quadratic is always positive, so the modulus may be dropped).
3. ∫ dx/x³ = ∫ x⁻³ dx = x⁻²/(−2) + C = −1/(2x²) + C. Do not use the log rule unless the power of x in the denominator is exactly one.
4. ∫ tan x dx = ∫ sin x/cos x dx = −ln|cos x| + C = ln|sec x| + C.
5. ∫ ln x dx = x ln x − x + C (by parts, section 08).

Geometric: a curve from its slope

• If a curve y = f(x) passes through a known point and the slope of its tangent at any point is given, integrate the slope to get y = ∫ (slope) dx, then substitute the point to find C.
• Example: dy/dx = 2x − 3 and the curve passes through (2, 5). y = x² − 3x + C; 5 = 4 − 6 + C → C = 7, so y = x² − 3x + 7.
• The slope of the tangent is dy/dx; if instead the normal is given, use dy/dx = −1/(normal slope).

Physical: motion along a straight line

• Acceleration a = dv/dt, so v = ∫ a dt; velocity v = ds/dt, so s = ∫ v dt — each integration needs one initial condition (initial velocity, initial position).
• Example: a body starts from rest with a = 6t − 4. v = 3t² − 4t, s = t³ − 2t²; at t = 3, v = 15 m/s and s = 9 m.
• Work done by a variable force: W = ∫ F ds; and the change of momentum is ∫ F dt (see the vector-functions lesson).

Worked example (curve through two conditions)

1. A curve has second derivative d²y/dx² = 12x − 2 and passes through (0, 1) with tangent slope 4 there. dy/dx = 6x² − 2x + C₁ → C₁ = 4. y = 2x³ − x² + 4x + C₂ → C₂ = 1. y = 2x³ − x² + 4x + 1.

The differential of a function

• For y = f(x) the differential is dy = f′(x) dx: the differential of the function equals its derivative multiplied by the differential of the independent variable.
• dx is the increment of x; dy is the corresponding change along the tangent, not along the curve. For small increments dy ≈ Δy, and the error shrinks as Δx → 0.
• Second differential: d²y = f″(x) dx² (only when x is the independent variable, so that dx is constant).
• Rewriting a differential equation in differential form is what makes direct integration legal: dy/dx = g(x)h(y) → dy/h(y) = g(x) dx, then integrate both sides.

Worked examples

1. y = x³ ln x → dy = (3x² ln x + x²) dx.
2. y = sin(2x + 1) → dy = 2 cos(2x + 1) dx.
3. Approximation: a cube of side 5 cm grows by 0.1 cm. V = x³, dV = 3x² dx = 3(25)(0.1) = 7.5 cm³ (the true increase is 7.31 cm³).
4. dy/dx = x/y with y(0) = 3 → y dy = x dx → y²/2 = x²/2 + C → C = 9/2, so y² = x² + 9.

The idea

• It reverses the chain rule: if the integrand contains a function of g(x) together with a multiple of g′(x), put u = g(x) and du = g′(x) dx.
• Steps: ① choose u (usually the inner function, the denominator, or the expression under a root) ② compute du and solve for dx ③ rewrite the whole integral in u ④ integrate ⑤ return to x.
• For a definite integral, either convert the limits to u or substitute back before applying the original limits.

Two useful standard results

• ∫ f′(x)/f(x) dx = ln|f(x)| + C.
• ∫ [f(x)]ⁿ f′(x) dx = [f(x)]ⁿ⁺¹/(n + 1) + C, n ≠ −1.

Worked examples

1. ∫ 2x(x² + 1)⁵ dx: u = x² + 1, du = 2x dx → ∫ u⁵ du = u⁶/6 + C = (x² + 1)⁶/6 + C.
2. ∫ cos⁴x sin x dx: u = cos x, du = −sin x dx → −∫ u⁴ du = −cos⁵x/5 + C.
3. ∫ dx/(x ln x): u = ln x, du = dx/x → ∫ du/u = ln|ln x| + C.
4. ∫ √(2x + 3) dx: u = 2x + 3, du = 2 dx → ½·(2/3)u^3/2 = (2x + 3)^3/2/3 + C.

The rule

• ∫ u dv = uv − ∫ v du, taken from the product rule d(uv) = u dv + v du.
• Choose u by LIATE: Logarithmic, Inverse trig, Algebraic (powers of x), Trigonometric, Exponential — the first type in this list becomes u; everything left, including dx, becomes dv.
• Use it when the integrand is a product of two unlike factors, e.g. x e^x, x sin x, x ln x, or a single logarithm or inverse trig function.

Worked examples

1. ∫ x e^x dx: u = x, dv = e^xdx → x e^x − ∫ e^x dx = x e^x − e^x + C.
2. ∫ x cos x dx: u = x, dv = cos x dx → x sin x − ∫ sin x dx = x sin x + cos x + C.
3. ∫ ln x dx: u = ln x, dv = dx → x ln x − ∫ x·(1/x) dx = x ln x − x + C.
4. ∫ x² e^x dx needs parts twice: x²e^x − 2x e^x + 2e^x + C.
5. Choosing the wrong u (for instance u = e^x in (1)) makes the new integral harder — always test whether ∫ v du is simpler than the original.

Choosing the method

1. ∫ (4x³ − 2x + 7) dx = x⁴ − x² + 7x + C
2. ∫ sin(3x + 1) dx = −cos(3x + 1)/3 + C
3. ∫ sec²(4x) dx = tan(4x)/4 + C
4. ∫ e^−x/2 dx = −2e^−x/2 + C
5. ∫ dx/(5 − 2x) = −ln|5 − 2x|/2 + C
6. ∫ cosec x cot x dx = −cosec x + C
7. ∫ cot x dx = ln|sin x| + C (compare with ∫ tan x dx = ln|sec x| + C)
8. ∫ x e^2x dx = (x e^2x)/2 − e^2x/4 + C
9. A curve with dy/dx = 3x² − 2 through (1, 0): y = x³ − 2x + 1
10. d y for y = x² sin x: (2x sin x + x² cos x) dx