# 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).