cos xdx Hence sin x = eu + c e eu du  c   sin x Example 5 10 Let u = tan x, then du = sec2 xdx ∫ Hence (tan x )10 sec2 xdx u10 du  u11  11  c tan11 x  11  c 4. Adjustment of the derivative Sometimes the integrand may not have the exact derivative of the function u present. In such cases we shall adjust the derivative so that the function u and its exact derivative u1 are present in the integrand. The following examples shall illustrate this principle. Example 6 2x3 3x3 6x 5    Find dx x  x  1 2 Solution The integrand has one function u = 2x3 + 3x2 + 6x + 5 Then u1 = 6x2 + 6x + 6 = 6 (x2 + x + 10) The other function present in the integrand is (x2 + x + 1). We must multiply this second function by six and divide by six to get u1. x x 1 1 6x 6x 1     Thus 2 2 Now the integrand has u and its derivative u1. Let u = 2x3 + 3x2 + 6x + 5 28 du dx
Hence e sin x cos xdx = eu + c  eu du  sin x  c Example 5 10 Let u = tan x, then du = sec2 xdx Hence ∫ (tan x )10 sec2 xdx  u10 du  u11 11  c  tan11 x 11  c 4. Adjustment of the derivative Sometimes the integrand may not have the exact derivative of the function u present. In such cases we shall adjust the derivative so that the function u and its exact derivative u1 are present in the integrand. The following examples shall illustrate this principle. Example 6 Find x 2  x  1 2x3  3x3  6x  5 dx Solution The integrand has one function u = 2x3 + 3x2 + 6x + 5 Then u1 = 6x2 + 6x + 6 = 6 (x2 + x + 10) The other function present in the integrand is (x2 + x + 1). We must multiply this second function by six and divide by six to get u1. Thus x 2  x  1 1 6x 2  6x  1 Now the integrand has u and its derivative u1. Let u = 2x3 + 3x2 + 6x + 5 28 du dx
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