Integral Generalizations

Just keeping a note to self that I can revise and study for my upcoming integration bee. I'd love your advice for studying for it here!

∫xnex dx=n!ex(xnn!−xn−1(n−1)!+xn−2(n−2)!−xn−3(n−3)!+⋯+)+C\displaystyle \int x^ne^x \ dx = n!e^x \left( \frac{x^n}{n!} - \frac{x^{n - 1}}{(n - 1)!} + \frac{x^{n - 2}}{(n - 2)!} - \frac{x^{n - 3}}{(n - 3)!} + \cdots + \right) + C

∫1(x+a)(x+b) dx=1a−bln⁡∣x+bx+a∣+C,a>b\displaystyle \begin{aligned} \int \frac{1}{(x + a)(x + b)} \ dx = \frac{1}{a - b} \ln \left| \frac{x + b}{x + a} \right| + C, & & a > b \end{aligned}

Ia=∫1xa+x dx=−1a−1ln⁡(1+1xa−1)+C\displaystyle I_a = \int \frac{1}{x^a + x} \ dx = -\frac{1}{a - 1}\ln \left(1 + \frac{1}{x^{a-1}}\right) + C

In=∫0π2sin⁡n(x) dx=∫0π2cos⁡n(x) dx,n≥3\displaystyle \begin{aligned} I_n = \int_0^{\frac{\pi}{2}} \sin ^n (x) \ dx = \int_0^{\frac{\pi}{2}} \cos ^n (x) \ dx, & & n \geq 3 \end{aligned}

  • I2k=12×34×56×⋯×2k−12k×π2  ⟸  12×⋯×one lessthe power itself×π2\displaystyle I_{2k} = \frac 12 \times \frac 34 \times \frac 56 \times \cdots \times \frac{2k - 1}{2k} \times \frac{\pi}{2} \ \ \Longleftarrow \ \ \frac 12 \times \cdots \times \frac{\small \text{one less}}{\small \text{the power itself}} \times \frac{\pi}{2}
  • I2k+1=23×45×67×⋯×2k2k+1       ⟸  23×⋯×one lessthe power itself\displaystyle I_{2k + 1} = \frac 23 \times \frac 45 \times \frac 67 \times \cdots \times \frac{2k}{2k + 1} \ \ \ \ \ \ \ \Longleftarrow \ \ \frac 23 \times \cdots \times \frac{\small \text{one less}}{\small \text{the power itself}}

Ia=∫01xa−1ln⁡x dx=ln⁡(a+1)+C\displaystyle I_a = \int_0^1 \frac{x^a - 1}{\ln x} \ dx = \ln (a + 1) + C

∫ex(f(x)+f′(x)) dx=exf(x)+C\displaystyle \int e^x \bigg( f(x) + f'(x) \bigg) \ dx = e^xf(x) + C

∫01xm(1−x)n dx=m!n!(m+n+1)!+C\displaystyle \int_0^1 x^m(1 - x)^n \ dx = \frac{m!n!}{(m + n + 1)!} + C

#Calculus

Note by Zach Abueg
3 years, 9 months ago

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