Calculus II - Integration Techniques

23 flashcards · Mathematics · by luckydog789 (harrison holbrook)

1. U-Substitution

   An integration technique which reverses the chain rule. Importantly, not all integrals yield to U-Substitution, as not all integrals implement the chain rule in their integrand.

2. Steps for U-Substitution

   Let u=g(x)u=g(x)u=g(x) , where ggg is differentiable on an interval, and let fff be continuous on the corresponding range of ggg . On that interval, ∫f(g(x))g′(x)dx=∫f(u)du\int f(g(x))g'(x)dx=\int f(u)du∫f(g(x))g′(x)dx=∫f(u)du Given an indefinite integral involving a composite function f(g(x))f(g(x))f(g(x)) , identify an inner function u=g(x)u=g(x)u=g(x) such that a constant multiple of g′(x)g'(x)g′(x) appears in the integrand Substitute u=g(x)u=g(x)u=g(x) and du=g′(x)dxdu=g'(x)dxdu=g′(x)dx in…

3. Integration by parts

   An integration technique that reverses the product rule. It produces an integral that can be easier to antidifferentiate than the original, and can be performed multiple times.

4. Integration by parts formula

   ∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx Let u=f(x),v=g(x)u=f(x), v=g(x)u=f(x),v=g(x) indefinite: ∫u dv=uv−∫v du\int u \text{ }dv=uv-\int v \text{ }du∫u dv=uv−∫v du definite: ∫abu dv=[uv]ab−∫abv du\int_a^b u \text{ } dv = [uv]_a^b- \int_a^bv \text{ }du∫ab​u dv=[uv]ab​−∫ab​v du

5. Steps for integration by parts

   Let u=f(x),v=g(x)u=f(x), v=g(x)u=f(x),v=g(x) Choose uuu and dvdvdv such that ∫u dv\int u \text{ } dv ∫u dv equals the desired integral Guidelines for choosing uuu and dvdvdv dxdxdx is always part of dvdvdv dvdvdv should be easy to integrate uuu and dvdvdv are chosen so that ∫v du\int v \text{ } du∫v du is no more difficult to integrate than the original integral ∫u dv\int u \text{ } dv∫u dv If the new integral is more complicated, try different choices for uuu and dvdvdv Find vvv by integrating…

6. Integrals of form ∫sin⁡n(x)dx\int\sin^n(x)dx∫sinn(x)dx , where 2"> n>2n>2n>2 is an integer and nnn is odd

   integral: ∫sin⁡n(x)dx=∫sin⁡n−1(x)sin⁡(x)dx\int\sin^n(x)dx=\int\sin^{n-1}(x)\sin(x)dx∫sinn(x)dx=∫sinn−1(x)sin(x)dx identity: sin⁡2(x)=1−cos⁡2(x)\sin^2(x)=1-\cos^2(x)sin2(x)=1−cos2(x) U-Sub: u=cos⁡(x)u=\cos(x)u=cos(x) , du=−sin⁡(x)dxdu=-\sin(x)dxdu=−sin(x)dx

7. Integrals of form ∫cos⁡n(x)dx\int\cos^n(x)dx∫cosn(x)dx , where 2"> n>2n>2n>2 is an integer and nnn is odd

   integral: ∫cos⁡n(x)dx=∫cos⁡n−1(x)cos⁡(x)dx\int\cos^n(x)dx=\int\cos^{n-1}(x)\cos(x)dx∫cosn(x)dx=∫cosn−1(x)cos(x)dx identity: cos⁡2(x)=1−sin⁡2(x)\cos^2(x)=1-\sin^2(x)cos2(x)=1−sin2(x) U-Sub: u=sin⁡(x)u=\sin(x)u=sin(x) , du=cos⁡(x)dxdu=\cos(x)dxdu=cos(x)dx

8. Integrals of form ∫sin⁡n(x)dx\int\sin^n(x)dx∫sinn(x)dx , where 2"> n>2n>2n>2 is an integer and nnn is even

   integral: ∫sin⁡2(x)dx=∫(1−cos⁡(2x)2)n/2dx\int\sin^2(x)dx=\int(\frac{1-\cos(2x)}{2})^{n/2}dx∫sin2(x)dx=∫(21−cos(2x)​)n/2dx identity: sin⁡2(x)=1−cos(2x)2\sin^2(x)=\frac{1-cos(2x)}{2}sin2(x)=21−cos(2x)​ U-sub: u=2xu=2xu=2x , du=2dxdu=2dxdu=2dx

9. Integrals of form ∫cos⁡n(x)dx\int\cos^n(x)dx∫cosn(x)dx , where 2"> n>2n>2n>2 is an integer and nnn is even

   integral: ∫cos⁡2(x)dx=∫(1+cos⁡(2x)2)n/2dx\int\cos^2(x)dx=\int(\frac{1+\cos(2x)}{2})^{n/2}dx∫cos2(x)dx=∫(21+cos(2x)​)n/2dx identity: cos⁡2(x)=1+cos(2x)2\cos^2(x)=\frac{1+cos(2x)}{2}cos2(x)=21+cos(2x)​ U-sub: u=2xu=2xu=2x , du=2dxdu=2dxdu=2dx

10. Integrals of the form ∫sin⁡m(x)cos⁡n(x)dx\int\sin^m(x)\cos^n(x)dx∫sinm(x)cosn(x)dx where mmm is odd

    integral: ∫sin⁡m(x)cos⁡n(x)dx=∫sin⁡m−1(x)cos⁡n(x)sin⁡(x)dx\int\sin^m(x)\cos^n(x)dx=\int\sin^{m-1}(x)\cos^n(x)\sin(x)dx∫sinm(x)cosn(x)dx=∫sinm−1(x)cosn(x)sin(x)dx identity: sin⁡2(x)=1−cos⁡2(x)\sin^2(x)=1-\cos^2(x)sin2(x)=1−cos2(x) U-Sub: u=cos⁡(x)u=\cos(x)u=cos(x) , du=−sin⁡(x)dxdu=-\sin(x)dxdu=−sin(x)dx

11. Integrals of the form ∫sin⁡m(x)cos⁡n(x)dx\int\sin^m(x)\cos^n(x)dx∫sinm(x)cosn(x)dx where nnn is odd

    integral: ∫sin⁡m(x)cos⁡n(x)dx=∫sin⁡m(x)cos⁡n−1(x)cos⁡(x)dx\int\sin^m(x)\cos^n(x)dx=\int\sin^{m}(x)\cos^{n-1}(x)\cos(x)dx∫sinm(x)cosn(x)dx=∫sinm(x)cosn−1(x)cos(x)dx identity: cos⁡2(x)=1−sin⁡2(x)\cos^2(x)=1-\sin^2(x)cos2(x)=1−sin2(x) U-Sub: u=sin⁡(x)u=\sin(x)u=sin(x) , du=cos⁡(x)dxdu=\cos(x)dxdu=cos(x)dx

12. Integrals of the form ∫sin⁡m(x)cos⁡n(x)dx\int\sin^m(x)\cos^n(x)dx∫sinm(x)cosn(x)dx where both mmm and nnn are even

    integral: ∫sin⁡m(x)cos⁡n(x)dx=∫sin⁡m(x)(1−sin2(x))n/2dx\int\sin^m(x)\cos^n(x)dx=\int\sin^m(x)(1-sin^2(x))^{n/2}dx∫sinm(x)cosn(x)dx=∫sinm(x)(1−sin2(x))n/2dx or ∫sin⁡m(x)cos⁡n(x)dx=∫(1−cos⁡2(x))m/2cos⁡n(x)dx\int\sin^m(x)\cos^n(x)dx=\int(1-\cos^2(x))^{m/2}\cos^n(x)dx∫sinm(x)cosn(x)dx=∫(1−cos2(x))m/2cosn(x)dx identity: sin⁡2(x)=1−cos(2x)2\sin^2(x)=\frac{1-cos(2x)}{2}sin2(x)=21−cos(2x)​ or cos⁡2(x)=1+cos(2x)2\cos^2(x)=\frac{1+cos(2x)}{2}cos2(x)=21+cos(2x)​ U-sub: u=2xu=2xu=2x , du=2dxdu=2dxdu=2dx

13. Integrals in the form ∫tan⁡m(x)sec⁡n(x)dx\int\tan^m(x)\sec^n(x)dx∫tanm(x)secn(x)dx where mmm is odd

    integral: ∫tan⁡m(x)sec⁡n(x)dx=∫tan⁡m−1sec⁡n−1sec⁡(x)tan⁡(x)dx\int\tan^m(x)\sec^n(x)dx=\int\tan^{m-1}\sec^{n-1}\sec(x)\tan(x)dx∫tanm(x)secn(x)dx=∫tanm−1secn−1sec(x)tan(x)dx identity: tan⁡2(x)=sec⁡2(x)−1\tan^2(x)=\sec^2(x)-1tan2(x)=sec2(x)−1 U-sub: u=sec⁡(x)u=\sec(x)u=sec(x) , du=sec⁡(x)tan⁡(x)dxdu=\sec(x)\tan(x)dxdu=sec(x)tan(x)dx

14. Integrals in the form ∫tan⁡m(x)sec⁡n(x)dx\int\tan^m(x)\sec^n(x)dx∫tanm(x)secn(x)dx where nnn is even

    integral: ∫tan⁡m(x)sec⁡n(x)dx=∫tan⁡m(x)sec⁡n−2(x)sec⁡2(x)dx\int\tan^m(x)\sec^n(x)dx=\int\tan^{m}(x)\sec^{n-2}(x)\sec^2(x)dx∫tanm(x)secn(x)dx=∫tanm(x)secn−2(x)sec2(x)dx identity: sec⁡2(x)=1+tan⁡2(x)\sec^2(x)=1+\tan^2(x)sec2(x)=1+tan2(x) U-sub: u=tan⁡(x)u=\tan(x)u=tan(x) , du=sec2(x)dxdu=sec^2(x)dxdu=sec2(x)dx

15. Integrals in the form ∫tan⁡m(x)sec⁡n(x)dx\int\tan^m(x)\sec^n(x)dx∫tanm(x)secn(x)dx where mmm is even and nnn is odd

    integral: ∫tan⁡m(x)sec⁡n(x)dx=∫(sec⁡2(x)−1)m/2sec⁡n(x)dx\int\tan^m(x)\sec^n(x)dx=\int(\sec^2(x)-1)^{m/2}\sec^n(x)dx∫tanm(x)secn(x)dx=∫(sec2(x)−1)m/2secn(x)dx identity: tan⁡2(x)=sec⁡2(x)−1\tan^2(x)=\sec^2(x)-1tan2(x)=sec2(x)−1 Method: Integration by Parts

16. Trig Sub

    Find the substitution Recalculate the bounds if necessary by setting the old bound equal to the substitution, then solving for θ\thetaθ Solve the integral If Indefinite: Use the trigonometric identities (triangle trig) to express this integral from terms of θ\thetaθ to terms of the original variable If definite: just evaluate with the recalculated bounds

17. a2−x2\sqrt{a^2-x^2}a2−x2 ​

    integrand contains: a2−x2\sqrt{a^2-x^2}a2−x2 ​ substitution: x=asin⁡(θ)x=a\sin(\theta)x=asin(θ) , −π2≤θ≤π2-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}−2π​≤θ≤2π​ based on the identity: 1−sin⁡2θ=cos⁡2θ1-\sin^2\theta=\cos^2\theta1−sin2θ=cos2θ

18. x2+a2\sqrt{x^2+a^2}x2+a2 ​

    integrand contains: x2+a2\sqrt{x^2+a^2}x2+a2 ​ substitution: x=atan⁡(θ)x=a\tan(\theta)x=atan(θ) , −π2≤θ≤π2-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}−2π​≤θ≤2π​ based on the identity: tan⁡2θ+1=sec⁡2θ\tan^2\theta+1=\sec^2\thetatan2θ+1=sec2θ

19. x2−a2\sqrt{x^2-a^2}x2−a2 ​

    integrand contains: x2−a2\sqrt{x^2-a^2}x2−a2 ​ substitution: x=asec⁡(θ)x=a\sec(\theta)x=asec(θ) , 0≤θ≤π20\leq\theta\leq\frac{\pi}{2}0≤θ≤2π​ , π≤θ≤3π2\pi\leq\theta\leq\frac{3\pi}{2}π≤θ≤23π​ based on the identity: sec⁡2θ−1=tan⁡2θ\sec^2\theta-1=\tan^2\thetasec2θ−1=tan2θ

20. Integrating quadratics via completing the square

    Factor the quadratic by completing the square Use the substitution u=x+b2au=x+\frac{b}{2a}u=x+2ab​ and du=dxdu=dxdu=dx , expressing the original quadratic as au2+rau^2+rau2+r , where r=c−b24ar=c-\frac{b^2}{4a}r=c−4ab2​ Solve

21. Integration of an improper* rational function via Long Division *(degree of the numerator is greater than that of the denominator)

    Use polynomial long division to express the integrand as a sum Hopefully, the remainder is easy to integrate Solve

22. LIATE (Integration by Parts)

    The general (not rigorous or strict) order in which to choose which terms to integrate vs differentiate in integration by parts. The term closer to L is typically differentiated, and the term closer to E is typically integrated. L - Logarithm I - Inverse Trig A - Algebraic T - Trigonometric E - Exponential

23. DI (Tabular) Method for Integration by Parts

    First, set up a table that looks like the following: \begin{matrix} & D & I \\ + & & \\ - & & \\ + & & \\ - & & \\ + & & \end{matrix}