Trigonometric Integrals Lecture Notes

Strategies for Integrating Powers of Sine and Cosine

• General Context: The focus is on evaluating integrals of the form $\int \sin^K(x) \cos^J(x) \, dx$. The strategy depends primarily on whether the powers of sine ($K$) or cosine ($J$) are odd or even.

• Case 1: The Power of Sine ($K$) is Odd

  • Procedure: Reduce the power of sine by one ($\sin^{K-1}(x)$), so there is a single $\sin(x)$ factor left over to act as the derivative for substitution.

  • Substitution: Let $u = \cos(x)$. Then $du = -\sin(x) \, dx$.

  • Identity: Convert the remaining even powers of sine into terms of cosine using the Pythagorean identity: $\sin^2(x) = 1 - \cos^2(x)$.

• Case 2: The Power of Cosine ($J$) is Odd

  • Procedure: Reduce the power of cosine by one ($\cos^{J-1}(x)$), peeling off one $\cos(x)$ to serve as the derivative for substitution.

  • Substitution: Let $u = \sin(x)$. Then $du = \cos(x) \, dx$.

  • Identity: Convert the remaining even powers of cosine into terms of sine using the Pythagorean identity: $\cos^2(x) = 1 - \sin^2(x)$.

Basic Example: Odd Power of Sine

• Problem: $\int \sin(x) \cos^4(x) \, dx$

• Analysis: Sine is to the power of 1, which is odd. Since there is only one sine, there is no need to reduce the power or apply identities.

• Execution:

  • Let $u = \cos(x)$.

  • Then $du = -\sin(x) \, dx$, which implies $-du = \sin(x) \, dx$.

  • Rewrite the integral: $\int -u^4 \, du$.

  • Antiderivative: $- \frac{u^5}{5} + C$.

• Final Substitution: $- \frac{\cos^5(x)}{5} + C$.

Advanced Example: Odd Power of Cosine

• Problem: $\int \sin^2(x) \cos^3(x) \, dx$

• Analysis: Cosine has the odd power ($3$). We peel off one cosine factor.

• Execution:

  • Rewrite: $\int \sin^2(x) \cos^2(x) \cos(x) \, dx$.

  • Convert $\cos^2(x)$: $\int \sin^2(x) (1 - \sin^2(x)) \cos(x) \, dx$.

  • Let $u = \sin(x)$. Then $du = \cos(x) \, dx$.

  • Substitute: $\int u^2 (1 - u^2) \, du = \int (u^2 - u^4) \, du$.

  • Integrate: $\frac{u^3}{3} - \frac{u^5}{5} + C$.

• Final Substitution: $\frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} + C$.

• Verification Note: If you check the work by taking the derivative, the result may look different from the original integrand. You must apply trigonometric identities (like factoring out $\sin^2(x)$ and converting $(1 - \sin^2(x))$ back to $\cos^2(x)$) to match the starting function.

Strategies for Even Powers (Double Angle Formulas)

• Context: When both sine and cosine have even powers, $-u$ substitution is not immediately possible because no single $\sin(x)$ or $\cos(x)$ is available for $du$. We must use power-reduction/double-angle identities.

• Core Identities:

  • $\sin^2(x) = \frac{1 - \cos(2x)}{2}$

  • $\cos^2(x) = \frac{1 + \cos(2x)}{2}$

• Example: Integrating $\sin^2(x)$:

  • $\int \sin^2(x) \, dx = \int \frac{1 - \cos(2x)}{2} \, dx = \int (\frac{1}{2} - \frac{\cos(2x)}{2}) \, dx$.

  • Integration: $\frac{1}{2}x - \frac{\sin(2x)}{4} + C$.

  • Note: The division by 4 comes from the substitution $u = 2x$, where $du = 2 \, dx$, adding another factor of $\frac{1}{2}$.

• Example: Integrating $\cos^2(x)$:

  • $\int \cos^2(x) \, dx = \int \frac{1 + \cos(2x)}{2} \, dx = \frac{1}{2}x + \frac{\sin(2x)}{4} + C$.

Scaling Factors in the Argument

• If the argument of the function is already scaled, such as $\sin^2(7x)$, the double angle formula doubles that specific argument.

• Example: $\int \sin^2(7x) \, dx$

  • Formula: $\frac{1 - \cos(2 \times 7x)}{2} = \frac{1 - \cos(14x)}{2}$.

  • Integration results in: $\frac{1}{2}x - \frac{\sin(14x)}{28} + C$.

Higher Even Powers: $\cos^4(x)$

• Strategy: Treat $\cos^4(x)$ as $(\cos^2(x))^2$.

• Step-by-Step Procedure:

  1. Apply identity: $\int (\frac{1 + \cos(2x)}{2})^2 \, dx$.

  2. Expand the square: $\int \frac{1}{4}(1 + 2\cos(2x) + \cos^2(2x)) \, dx$.

  3. Distribute the constant: $\int (\frac{1}{4} + \frac{1}{2}\cos(2x) + \frac{1}{4}\cos^2(2x)) \, dx$.

  4. Apply double angle identity again to $\cos^2(2x)$: $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$.

  5. The third term becomes: $\frac{1}{4} (\frac{1 + \cos(4x)}{2}) = \frac{1}{8} + \frac{\cos(4x)}{8}$.

  6. Combine constants: $\frac{1}{4} + \frac{1}{8} = \frac{3}{8}$.

  7. Final Antiderivative: $\frac{3}{8}x + \frac{\sin(2x)}{4} + \frac{\sin(4x)}{32} + C$.

Integrals with Differing Scaling Factors (Product-to-Sum)

• When the scaling factors inside sine and cosine are different (e.g., $\sin(mx)\cos(nx)$), use product-to-sum identities derived from sum and difference formulas.

• Identity 2 (Sine and Cosine): $\sin(Ax)\cos(Bx) = \frac{1}{2}[\sin((A-B)x) + \sin((A+B)x)]$

• Example: $\int \sin(5x)\cos(3x) \, dx$

  • Apply identity: $\int \frac{1}{2}(\sin(2x) + \sin(8x)) \, dx$.

  • Integrate: $\frac{1}{2}(-\frac{\cos(2x)}{2} - \frac{\cos(8x)}{8}) + C$.

Strategies for Secant and Tangent Integrals

• Standard Integrals to Memorize:

  • $\int \sec^2(x) \, dx = \tan(x) + C$

  • $\int \sec(x)\tan(x) \, dx = \sec(x) + C$

  • $\int \tan(x) \, dx = \ln|\sec(x)| + C$

  • $\int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C$

• Pythagorean Identity: $\tan^2(x) + 1 = \sec^2(x)$.

• Case A: The Power of Secant ($J$) is Even ($J \ge 2$)

  • Goal: Use $u = \tan(x)$. Peel off a $\sec^2(x)$ for $du$.

  • Example: $\int \tan^6(x) \sec^4(x) \, dx$

    • Peel off $\sec^2(x)$: $\int \tan^6(x) \sec^2(x) \sec^2(x) \, dx$.

    • Convert one $\sec^2(x)$ to $\tan^2(x) + 1$.

    • Substitute $u = \tan(x), du = \sec^2(x) \, dx$.

    • Resulting integral: $\int u^6 (u^2 + 1) \, du = \int (u^8 + u^6) \, du$.

    • Answer: $\frac{\tan^9(x)}{9} + \frac{\tan^7(x)}{7} + C$.

• Case B: The Power of Tangent ($K$) is Odd ($K \ge 1$)

  • Goal: Use $u = \sec(x)$. Peel off a $\sec(x)\tan(x)$ for $du$.

  • Example: $\int \tan^3(x) \sec^3(x) \, dx$

    • Peel off: $\int \tan^2(x) \sec^2(x) (\sec(x)\tan(x)) \, dx$.

    • Convert $\tan^2(x)$ to $\sec^2(x) - 1$.

    • Substitute $u = \sec(x), du = \sec(x)\tan(x) \, dx$.

    • Integral: $\int (u^2 - 1)u^2 \, du = \int (u^4 - u^2) \, du$.

    • Answer: $\frac{\sec^5(x)}{5} - \frac{\sec^3(x)}{3} + C$.

• Case C: Pure Power of Tangent ($J = 0$)

  • Strategy: Separate $\tan^2(x)$ and convert it to $\sec^2(x) - 1$. Distribute and repeat the process if necessary.

• Case D: Pure Power of Secant (Integration by Parts)

  • Example: $\int \sec^3(x) \, dx$:

    • Use Integration by Parts ($\int u \, dv = uv - \int v \, du$).

    • Let $u = \sec(x)$ and $dv = \sec^2(x) \, dx$.

    • Then $du = \sec(x)\tan(x) \, dx$ and $v = \tan(x)$.

    • Apply IBP: $\sec(x)\tan(x) - \int \tan^2(x)\sec(x) \, dx$.

    • Convert $\tan^2(x)$: $\sec(x)\tan(x) - \int (\sec^2(x) - 1)\sec(x) \, dx$.

    • This results in: $\sec(x)\tan(x) - \int \sec^3(x) \, dx + \int \sec(x) \, dx$.

    • Recognize the "boomerang": Add $\int \sec^3(x) \, dx$ to both sides to get $2 \int \sec^3(x) \, dx$.

    • Solve: $\int \sec^3(x) \, dx = \frac{1}{2} [\sec(x)\tan(x) + \ln|\sec(x) + \tan(x)|] + C$.

Questions & Discussion

• Question: How did you go from $\int \sec^3(x) \dots$ to $\sec^5(x)$ in the explanation?

• Answer: In the derivation of higher powers, you combine $\sec^3(x)$ and $\sec^2(x)$. $\sec^3(x) \times \sec^2(x) = \sec^{3+2}(x) = \sec^5(x)$. (Correction of minor error in verbal transcription regarding power of 6).

• Discussion on Definite Integrals: Definite trigonometric integrals are evaluated the same way but require an extra step of algebra at the end to plug in bounds (e.g., $0$ to $\pi$). The instructor notes that these can be very long due to the algebra involved.

• Future Schedule: Thursday will finish up trig integrals, work on integration by parts from homework, and spend the last 20-25 minutes reviewing. Next week will likely include one trig problem and then definite integrals.