Integration by parts (product rule backwards) The product rule states Integrating both sides gives f(x)g(x) = d dx f(x)g(x) = f(x)g (x) + f (x)g(x). f(x)g (x)dx + Letting f(x) = u, g(x) = v, and rearranging, we obtain u dv = uv v du. Some examples: f (x)g(x)dx. xe x dx. Differentiating u = x gives a simpler function, du = dx, while integrating dv = e x dx gives v = e x which is no more difficult to integrate. Applying integration by parts gives xe x dx = xe x e x dx = xe x e x = e x (x ). ln x dx. Differentiating u = ln x gives the simpler function du = xdx while integrating dv = dx gives v = x. Applying integration by parts, we obtain ln x dx = x ln x x dx = x ln x x = x(ln x ). x e x sin x dx. We can integrate by parts twice and solve algebraically for the integral as follows: ( ) I = e x sin x dx = e x sin x e x cos x dx = e x sin x e x cos x e x ( sin x)dx = e x (sin x cos x) e x sin x dx = e x (sin x cos x) I, I = e x (sin x cos x), I = ex (sin x cos x).
Some for you to try: ln y. dy y u = ln y dv = y / dy du = dy/y v = y / ln y dy = y / ln y y y / dy = y / ln y 4y / = y(ln y ).. x e x dx 3. x dx = ex x ln x dx xe x dx = xe x u = x dv = e x dx du = dx v = e x / + e x dx = xe x e x 4 = e x (x /). x ln x dx = x ln x u = ln x dv = xdx du = dx/x v = x / x dx = x x ln x 4 = x (ln x /) 4. arccos z dz u = arccos z du = dz z dv = dz v = z arccos z dz = z arccos z + z dz. z We now make the substitution u = z, du = zdz to obtain z dz = u / du = u z =. Hence arccos z dz = z arccos z + z = + =. z
5. x 3 + x dx You don t need integration by parts. the substitution x 3 + x dx = (u )du = u u = + x, du = xdx, x = u, = 3 (x ) + x. However, you could use integration parts with (u / u / )du = 3 u3/ u / = (u/3 ) u u = x dv = x +x du = xdx v = + x (using a substitution to find dv). We get (with w = + x, dw = xdx) x 3 + x dx = x + x x + x = x w + x dw = x + x 3 u3/ = x + x 3 ( + x ) 3/ = 3 (x ) + x. 6. 3 arctan(/x)dx 3 u = arctan(/x) du = dx x + dv = dx v = x arctan(/x)dx = x arctan(/x) w = + x, dw = xdx, the resulting integral is so that the end result is 3 3 x dx + x = 4 dw w arctan(/x)dx = x arctan(/x) 3 3 3 x dx + + x. = ln 4 ln = ln, + ln = 3 π 6 π 4 + ln = ( 3 3)π + ln.
7. (ln x) dx (ln x) dx = x(ln x) u = (ln x) dv = dx du = ln x x dx v = x ln x dx = x(ln x) (x ln x x) = x(ln x) x ln x + x. 8. t sin(πt)dt We will use integration by parts twice to get rid of the t. u = t du = tdt dv = sin(πt)dt v = cos(πt)/π t sin(πt)dt = t π cos(πt) + π Integration by parts on the resulting integral with t cos(πt)dt. u = t du = dt dv = cos(πt)dt v = sin(πt)/π 9.. gives Hence e x cos(πx)dx t cos(πt)dt = t π sin(πt) π t sin(πt)dt = t π sin(πt)dt = t π sin(πt) + cos(πt) π. t cos(πt) cos(πt) + sin(πt) + π π 3. This is similar to the third example (on the first page). cos( x)dx (Make a substitution first.) the substitution y = x, dy = dx, the integral becomes x cos( x)dx = y cos(y)dy. Now use integration by parts with u = y dv = cos y dy du = dy v = sin y.
. sin(ln w)dw (Make a substitution first.) the substitution x = ln w, dx = dw/w, the integral becomes sin(ln w)dw = e x sin x dx.. Now integrate as in the third example (on the first page). sec 3 θ dθ the integral becomes u = sec θ du = sec θ tan θ sec 3 θ dθ = sec θ tan θ dv = sec θdθ v = tan θ tan θ sec θ dθ. Writing tan θ = sec θ gives tan θ sec θ dθ = (sec θ ) sec θ dθ = sec 3 θ dθ sec θ dθ = sec 3 θ dθ ln sec θ+tan θ. Hence I = sec 3 θ dθ = sec θ tan θ sec 3 θ dθ + ln sec θ + tan θ, I = sec θ tan θ + ln sec θ + tan θ, I = (sec θ tan θ + ln sec θ + tan θ ).