1. -2 cos(x/2) + c

    1. sin(x³) / 3

    1. Substitute u = x³. This gives du = 3 x² dx, and we get ∫ x² cos(x3) dx = ∫ cos(u) / 3 du = sin(u) /          3 + c = sin(x³)/3

    1. ln ( 1 + x2 ) / 2 + c

    1. -ln[ cos(x) ] + c

    1. tan(x) = sin(x) / cos(x). Here we substitute u = cos(x), which means that du = – sin(x) dx, and            we get ∫ sin(x) / cos(x) dx = ∫ – 1 / u du = – ln(u) + c = – ln [ cos(x) ] + c Note: this is often listed          as-ln [ cos(x) ] + c = ln [ 1 / cos(x) ] + c = ln [ sec(x) ] + c

    1. – 1/4 cos⁴(x) + c

    1. Substitute u = cos(x), then du = – sin(x), and we have ∫ cos³(x) sin(x) dx = ∫ – u³ du = – 1/4 u⁴ +          c = – 1/4 cos⁴(x) + c