# Recitation 2

Definition 1: Suppose $f$ is a function. It is even if $f(-x)=f(x)$; it is odd if $f(-x)=-f(x)$.

Example 2: $f(x) = 1 - e^{-x^2}$ is even, whereas $g(x) = \arctan x$ is odd.

Theorem 3: Suppose $f$ is continuous on $[-a, a]$. (a) If $f$ is even, then $\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx$. (b) If $f$ is odd, then $\int_{-a}^a f(x)dx=0$.

Example 4: Evaluate (a) $\int e^u du$; (b) $\int ue^u du$; (c) $\int u^2e^u du$.

Answer: (a) $e^u+c$; (b) $(u-1)e^u$; (c) $(u^2-2u+2)e^u$.

Problem 5: Evaluate $\int (\ln x)^2 dx$.

Hint: Let $u = \ln x$.

Problem 6: Evaluate $\int z^3e^z dz$.

Hint: $\int z^3e^z dz = \int z^3 de^z = z^3e^3 - 3\int z^2e^zdz$.

Problem 7: Evaluate $\int_0^1 (x^2+1)e^{-x} dx$.

Hint: Let $u = -x$.

Problem 8: Evaluate $\int_0^\pi e^{\cos t}\sin 2tdt$.

Hint: Let $u=\cos t$. $\int_0^\pi e^{\cos t}\sin 2tdt = \int_0^\pi e^{\cos t}2\sin t\cos tdt = 2\int_{-1}^1 e^uudu$.

Problem 9: Evaluate $\int x\ln (1+x)dx$.

Hint: Let $u = \ln (1+x)$.

Problem 10: $\int e^{-3\theta}\cos 5\theta d\theta$.

Solution: The standard solution uses integral by parts twice. However, here we present a solution using complex numbers. Notice that the answer is equal to the real part of $\int e^{-3\theta}(\cos 5\theta + i\sin 5\theta)d\theta = \int e^{(-3+5i)\theta}d\theta = \frac{1}{-3+5i}e^{(-3+5i)\theta}$, which is equal to $e^{-3\theta}\left(\frac{-3}{34}\cos 5\theta + \frac{5}{34}\sin 5\theta\right)$.