By the integration by parts, we haveZ π−πf (x) cos(nx)dx = 1n f (x) sin(nx)π−π− 1nZ π−πf (x) sin(nx)dx

Let f : [0, π] → R be defined byf (x) =(0 if x = 0,x sin 1x − 1x cos 1x if x̸ = 0.Is f continuous?

Suppose that f is a continuous function and that S1−1f (x)dx = π. What is the valueofSπ0f (cos(x)) sin(x)dx?A. 0B. -πC. πD. 2πE. The integral is undefined

Consider the following function: f(x) = { sin x , if 0 ≤ x ≤ π 4 cos x , if π 4 ≤ x ≤ π 2 . Expand f(x) in a Fourier series of sine terms.

Evaluate the integral. (Use C for the constant of integration.)(x2 + 2x) cos(x) dx

Using the same substitution u = sin(x) enables us to do16 sin4(x) cos(x) dx = 16u4 du.In terms of u, we get + C, which, in terms of sin(x), becomes + C.