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.

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.

Making this substitution using16 sin2(x) cos3(x) dxgives us 16 sin2(x) (1 − sin2(x)) cos(x) dx = 16 sin2(x) cos(x) dx −  cos(x)  dx.

he surface determined by the parametric equations x = z(cos u + u sin u), y = z(sin u − u cos u),0 ≤ u, z ≤ 1

Evaluate the following integrals.(b)∫ 21xe3x dx (c)∫ cos(√x)√x dx (d)∫ x2 + 3x + 3x + 1 dx

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