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

[1 - cos(nπ/2) - sin(nπ/2)]^2Z transform of it

sinn x dx = − 1n sinn−1 x · cos x + n − 1n∫sinn−2 x dx. (6.4)If we take dv = sin x dx, then we have v = − cos x and we may integrate by parts withu = sinn−1 x, du = (n − 1) sinn−2 x · cos x.Using the fact that sin2 x + cos2 x = 1, one may thus conclude that∫sinn x dx = − sinn−1 x · cos x + (n − 1)∫sinn−2 x · cos2 x dx= − sinn−1 x · cos x + (n − 1)∫sinn−2 x · (1 − sin2 x) dx= − sinn−1 x · cos x + (n − 1)∫sinn−2 x dx + (1 − n)∫sinn x dx.Here, the rightmost integral coincides with the original integral on the left. Once we nowrearrange terms, we end up with n copies of the integral and equation (6.4) follows. Example 6.11 We use a reduction formula to compute the integral I3 in the case thatIn =∫xne2x dx.If we take u = xn and dv = e2x dx, then du = nxn−1 dx and v = 12 e2x, so one hasIn = 12 xne2x − n2∫xn−1e2x dx = 12 xne2x − n2 · In−1. (6.5)We now apply the last formula repeatedly to determine I3. According to the formula,I3 = 12 x3e2x − 32 · I2 = 12 x3e2x − 32 ·[12 x2e2x − I1]= 12 x3e2x − 32 ·[12 x2e2x − 12 xe2x + 12 · I0]= 12 x3e2x − 34 x2e2x + 34 xe2x − 34∫e2x dx= 12 x3e2x − 34 x2e2x + 34 xe2x − 38 e2x + C. 

Prove that the function f (x) = cos(πx/2) + cos(πx) has a root in eachinterval [4k − 1, 4k] and [4k, 4k + 1], for every k ∈ Z. Relevant theoremsand properties of the cosine function may be freely used, if stated clearly

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

∫e sin(x) cos(x)dx.