Trig Integrals: U-substitution and Identities

U-substitution and Trig Identities: Study Notes

• Objective of today’s discussion

  • Use u-substitution to simplify integrals involving products of powers of sine and cosine.
  • Leverage fundamental trig identities to manipulate integrands into a form solvable by u-substitution.
  • Develop a general algorithm/pattern for these trig integrals and practice with progressively harder examples.

• Quick recap: key trig identity (Pythagorean) and derived forms

  • Core identity: \sin^2 x + \cos^2 x = 1
  • From the core identity, derive other useful relations by dividing by appropriate terms:
  • Dividing by \sin^2 x:
    • \frac{\sin^2 x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} = \frac{1}{\sin^2 x}
    • This yields 1 + \cot^2 x = \csc^2 x, with the explicit pieces:
    • \cot^2 x = \frac{\cos^2 x}{\sin^2 x}
    • \csc^2 x = \frac{1}{\sin^2 x}
  • Dividing by \cos^2 x:
    • \tan^2 x + 1 = \sec^2 x, with:
    • \tan^2 x = \frac{\sin^2 x}{\cos^2 x}
    • \sec^2 x = \frac{1}{\cos^2 x}
  • These identities are the “four commonly used trigger identities” that often come up in these problems:
  • \sin^2 x + \cos^2 x = 1
  • 1 + \tan^2 x = \sec^2 x
  • 1 + \cot^2 x = \csc^2 x
  • Practical use: knowing these helps rewrite parts of an integrand into terms that pair well with a substitution or a standard form.

• Core strategy (motivation and general pattern)

  • If the integrand has a mix of sine and cosine powers, try to pick the substitution based on an odd power:
  • If the sine power is odd, use u-sub with u = \cos x (since du = -\sin x\,dx).
  • If the cosine power is odd, use u-sub with u = \sin x (since du = \cos x\,dx).
  • The remaining even powers of the other trig function are rewritten using the Pythagorean identity, e.g., replace cos^2 x by 1 - sin^2 x or sin^2 x by 1 - cos^2 x, turning the integrand into a polynomial in the chosen u.
  • When both powers are even, consider power-reduction or double-angle identities to rewrite the integrand into a form solvable by standard techniques (often still leading to a polynomial in sin x or cos x after substitution).
  • The “one simple substitution” idea: reduce the problem to integrating a polynomial in the chosen variable, i.e., something of the form \int P(u)\,du, and then substitute back for the original trig function.
  • A useful mental model: the hardest single step often governs the difficulty (e.g., the simplest nontrivial substitution or a single application of a trig identity).

• Example 1 (basic pattern): (\int \cos^6 x \sin x\,dx)

  • Step 1: Recognize odd power of sine: use u = \cos x\,,\; du = -\sin x\,dx\;\Rightarrow\; -du = \sin x\,dx
  • Step 2: Use identity to rewrite the remaining sine powers in terms of cosine:
  • Rewrite sin x dx as part of the substitution and express sin^2 x as 1 − cos^2 x:
    • Integrand becomes \cos^6 x \cdot \sin x\,dx = -\cos^6 x (1 - \cos^2 x) \, du
  • Step 3: Substitute: let u = \cos x, so the integral becomes
  • -\int u^6(1 - u^2)\,du = -\int (u^6 - u^8)\,du
  • Evaluate: -\left( \frac{u^7}{7} - \frac{u^9}{9} \right) + C = -\frac{u^7}{7} + \frac{u^9}{9} + C
  • Step 4: Back-substitute u = \cos x:
  • \boxed{\; -\frac{\cos^7 x}{7} + \frac{\cos^9 x}{9} + C \;}
  • Observations from this example:
  • The middle factor (1 − cos^2 x) expanded to produce a polynomial in u, giving multiple terms after expansion, e.g., a term proportional to u^6 and a term proportional to u^8, leading to two antiderivative terms.
  • The overall negative sign comes from du = −sin x dx.

• Example 1 takeaway: the same u-substitution approach can turn a product of a high power of cosine and a single sine into a simple polynomial in u once the sine-power is factored and rewritten via the identity.

• Example 2 (slightly harder): when there are more factors and mixed odd/even powers

  • Scenario considered: an integrand with an odd cosine power and an even sine power, e.g., something like \cos^5 x \sin^4 x\,dx
  • Pattern used in the lecture:
  • Since the cosine power is odd, factor one cos x to pair with dx, and use substitution with the sine variable:
    • Let u = \sin x\,,\; du = \cos x\,dx
  • Express the remaining even powers of cosine in terms of sine using the identity \cos^2 x = 1 - \sin^2 x:
    • \cos^4 x = (\cos^2 x)^2 = (1 - \sin^2 x)^2 = 1 - 2u^2 + u^4
  • The integral becomes a polynomial in u:
    • \int \cos x \cdot \cos^4 x \cdot \sin^4 x\,dx = \int (1 - 2u^2 + u^4) \cdot u^4 \,du = \int (u^4 - 2u^6 + u^8) \,du
  • Step 2: Integrate each term:
  • \int (u^4 - 2u^6 + u^8) \,du = \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9} + C
  • Step 3: Back-substitute u = \sin x:
  • \boxed{\; \frac{\sin^5 x}{5} - \frac{2\sin^7 x}{7} + \frac{\sin^9 x}{9} + C \;}
  • Additional notes from the lecture about this example:
  • The approach shows how you can turn a product with an odd-power cosine into a polynomial in sin x, via the substitution depending on which function has the odd power.
  • There was discussion about expanding and possibly using other trig identities (e.g., double-angle or power-reduction) if needed to simplify remaining terms, but the core method remains a straightforward polynomial in the chosen variable after substitution.

• A second-level insight: using identities to handle even-power cases with double-angle concepts

  • In some problems, after substitution and expansion you encounter expressions like cos(2x) or cos(4x). A common move is to apply identities such as
  • \cos^2 t = \frac{1 + \cos(2t)}{2}
  • This can convert squared terms into linear combinations of constants and cosines of multiples of the angle, facilitating straightforward antiderivatives (e.g., integrals of constants and cos(kx)).
  • The lecture demonstrated a step where a term involving cos^2(2x) was rewritten to simplify the integration, recognizing that the inner-angle substitution doubles the angle argument when applying the identity.

• Special case: the tangent integral (a clean, one-variable substitution)

  • Problem form: \int \tan x\,dx = \int \frac{\sin x}{\cos x} \, dx
  • Quick substitution method:
  • Let u = \cos x,\; du = -\sin x\,dx
  • The integral becomes -\int \frac{1}{u} \, du = -\ln|u| + C = -\ln|\cos x| + C
  • Alternative form: \ln|\sec x| + C, since \sec x = 1/\cos x
  • Important caution: Do not casually split \int \frac{\sin x}{\cos x} \, dx into \int \sin x \, dx - \int \cos x \, dx; this is not a valid rule for this integrand. The correct approach is a single u-substitution as shown.

• Quick study tips and exam-oriented takeaways

  • Memorize the four trigger identities and know how to derive them quickly from sin^2 x + cos^2 x = 1:
  • 1 + \tan^2 x = \sec^2 x
  • 1 + \cot^2 x = \csc^2 x
  • For a trig integrand with powers of sine and cosine, identify the odd power first and perform the corresponding substitution (u = sin x or u = cos x).
  • Rewrite remaining even powers using the Pythagorean identities (e.g., sin^2 x = 1 − cos^2 x or cos^2 x = 1 − sin^2 x) to obtain a polynomial in the chosen u.
  • If both powers are even, be prepared to use power-reduction or double-angle formulas to reduce to a solvable form, often ending with a polynomial in sin x or cos x after substitution.
  • When encountering expressions like cos^2(2x) or higher multiples, consider using identities to rewrite in terms of cos(nx) and then simplify, keeping track of how the angle changes under substitution.
  • Always substitute back after integrating in u, and include the constant of integration: +C.

• Connections to broader concepts

  • The u-substitution approach to trig integrals mirrors the general chain-rule idea: identify an inner function whose differential appears in the integrand and substitute to simplify.
  • The pattern of choosing which trig function to substitute based on odd powers aligns with reducing the integrand to a polynomial, which is a familiar and manageable class of integrals.
  • Practice with these examples strengthens intuition for when to apply identities versus when to rely on a clean substitution, a skill that carries into more complex calculus problems.

• Quick reference summary (synthetic cheatsheet)

  • For integrals of the form ∫ cos^m x sin^n x dx:
  • If n is odd, use u = cos x (du = - sin x dx).
  • If m is odd, use u = sin x (du = cos x dx).
  • If both m and n are even, use power-reduction or identities to create a solvable polynomial in sin or cos.
  • Common results to remember:
  • If I = ∫ cos^6 x sin x dx, then with u = cos x: I = -\frac{u^7}{7} + \frac{u^9}{9} + C, back-substituting u = cos x yields -\frac{\cos^7 x}{7} + \frac{\cos^9 x}{9} + C.
  • If I = ∫ tan x dx, then I = -\ln|\cos x| + C = \ln|\sec x| + C.

• Final note on the speaking style from the lecture

  • The instructor emphasized that you should become comfortable with manipulating the integrand using identities to reach a solvable form rather than multitasking many different methods at once.
  • The focus is on building an algorithmic approach: identify odd powers, substitute, reduce, and then integrate, returning to x at the end.

• End of notes on today’s content

  • If you want extra practice, try constructing problems with the same patterns: (a) cos^even x sin^odd x, (b) cos^odd x sin^even x, (c) both even with power-reduction, and (d) a pure tan x case, and apply the substitution strategies and identities discussed here.