# Integration by Parts

 Tutorials with examples and detailed solutions and exercises with answers on how to use the technique of integration by parts to find integrals. Review: The method of integration by parts may be used to easily integrate products of functions. If u and v are both functions of x then u (dv/dx) dx = uv - (du/dx) v dx In what follows K is a constant of integration. Example 1: Evaluate the integral 3 x ex dx Solution to Example 1: Let u = x and dv/dx = ex, hence du/dx = 1 and v = ex and using the method of integration by parts, we obtain 3 x ex dx = 3 x ex dx = 3 ( x ex - 1 . ex dx) = 3 x ex - 3 ex + K Example 2: Evaluate the integral x sin(x) dx Solution to Example 2: Let u = x and dv/dx = sin(x), hence du/dx = 1 and v = - cos(x) x sin(x) dx = x (-cos(x)) - 1 . (-cos(x)) dx = - x cos(x) + sin(x) + K Example 3: Evaluate the integral x2 cos(x) dx Solution to Example 3: Let u = x2 and dv/dx = cos(x), hence du/dx = 2x and v = sin(x) x2 cos(x) dx = x2 sin(x) - 2x sin(x) dx We now need to apply the method of integration by parts to the integral 2x sin(x) dx to obtain the final integral. In example 2 we determined the integral x sin(x) dx and we may use that result. Hence x2 cos(x) dx = x2 sin(x) - 2 (- x cos(x) + sin(x)) + K = x2 sin(x) + 2xcos(x) - 2sin(x) + K Example 4: Evaluate the integral x ln(x) dx Solution to Example 4: Let u = ln(x) and dv/dx = x, hence du/dx = 1/x and v = x2 / 2 x2 ln(x) dx = (x2 / 2) ln(x) - (x2 / 2) (1 / x) dx = (x2 / 2) ln(x) - (1/4) x2 + K Example 5: Evaluate the integral x cos(x/3) dx Solution to Example 5: Let u = x and dv/dx = cos(x/3), hence du/dx = 1 and v = 3 sin(x/3)2 / 2 x cos(x/3) dx = x (3sin(x/3)) - 1 . 3 sin(x/3) dx = 3xsin(x/3) + 9 cos(x/3) + K Example 6: Use integration by parts to evaluate the integral ln(x) dx Solution to Example 6: We first rewrite ln(x) as 1 . ln(x), hence ln(x) dx = 1 . ln(x) dx Let u = ln(x) and dv/dx = 1, hence du/dx = 1/x and v = x. Using integration by parts, we obtain 1 . ln(x) dx = x ln(x) - x (1/x) dx = x ln(x) - x + K Example 7: Use integration by parts to evaluate the integral x2 (ln(x))2 dx Solution to Example 7: Let dv/dx = x2 and u = (ln(x))2 and use integration by parts as follows I = x2 (ln(x))2 dx = [x3 / 3 (ln(x))2] - (x3 / 3)(2 ln(x) / x) dx = [x3 / 3 (ln(x))2] - (2/3)x2 ln(x) dx We now let dv/dx = x2 and u = ln(x) and use the integration by parts one more time = [x3 / 3 (ln(x))2] - (2/3)[ (x3 / 3 ln(x) - (x3 / 3) (1/x) dx ] = [x3 / 3 (ln(x))2] - (2/9)(x3 ln(x) + (2/9) x2 dx Integrate the last term = [(ln(x))2 x3 / 3 ] - (2/9)(x3 ln(x) + (2/27)x3 + K Example 8: Use integration by parts to evaluate the integral ex sin(2x) dx Solution to Example 8: We first define I = ex sin(2x) dx, v = sin(2x) and du/dx = ex and use integration by parts as follows I = ex sin(2x) dx = [ex sin(2x)] - ex 2 cos(2x) dx We now let V = ex and du/dx = ex and use the integration by parts one more time = [ex sin(2x)] - 2[ex cos(2x)] + 2 ex 2 (- sin(2x)) dx = [ex sin(2x)] - 2[ex cos(2x)] - 4ex sin(2x) dx We can write the integral I as follows I = [ex sin(2x)] - 2[ex cos(2x)] - 4I and solve the last equation for I I = (1/5)(ex sin(2x) - 2ex cos(2x)) + K Exercises: Use the table of integrals and the method of integration by parts to find the integrals below. [Note that you may need to use the method of integration by parts more than once]. 1. x cos(x) dx 2. x e2x dx 3. x1/3 ln(x) dx 4. ln(x) / (x2) dx 5. x3 cos(x) dx 6. x2 e-3x dx 7. ex cos(2x) dx Answers to Above Exercises 1. x sin(x) + cos(x) + K 2. (x/2)e2x - (1/4)e2x + K 3. (3/4)x3/4 ln(x) - (9/16)x3/4 + K 4. -ln(x) / x - 1/x + K 5. 3(x2 - 2)cos(x) + (x3 - 6x)sin(x) + K 6. -(1/27)(9x2 + 6x + 2) e-3x dx + K 7. (1/5)(excos(2x) + 2 exsin(2x)) More references on integrals and their applications in calculus.