Integration by Parts

Integration by Parts Integration by parts is a method we can use to evaluate integrals like Z f (x)g(x)dx where the term f (x) is easy to differentiate, and g(x) is easy to integrate. Integration by parts allows us to evaluate the integral in terms of the derivative of f (x) and the integral of g(x). An example of an integral we can solve using integration by parts is Z xex dx because x is easily differentiable, and ex is easily integrable. The product rule for derivatives states that d dv du u(x)v(x) = u(x) + v(x) dx dx dx If we integrate both sides of the equation we find Z Z Z d dv du u(x)v(x) = u(x) + v(x) dx dx dx which implies dv u(x) + dx Z du v(x) = u(x)v(x) − dx Z Z u(x)v(x) = v(x) du dx u(x) dv dx finally we rewrite Z If we know how to differentiate v(x), and how to integrate u(x), then we can evaluate the above integral. We can also restate the above formula in terms of differentials Z Z vdu = uv − udv where u,v are functions of x. The benefit of this form is that it is more compact and easier to remember than the Zprevious. Example 1 Evaluate xex dx Solution The first step to integrate is to choose which term will be used for u(x) and which dv will be dx . We make the choices u(x) = x and dv = ex dx as these functions are easy to differentiate and integrate respectively. When we integrate ex we will set the arbitrary constant to 0 for convenience. Thus, we find that du =1 dx and v(x) = ex Using the formula for integration by parts Z Z x x xe dx = xe − 1 · ex dx = xex − ex + c Z Example 2 Evaluate ln(x)dx Solution In this case we can write ln(x) = 1 · ln(x) and apply integration by parts. Since we don’t know how to integrate ln(x), we will set v(x) = ln(x) and We find du =1 dx dv 1 = dx x and u(x) = x once again setting the arbitrary constant to 0 (which we will always do with integration by parts). Finally Z Z 1 ln(x)dx = x · ln(x) − x dx = x ln(x) − x + c x Z Example 3 Evaluate x sec(x)2 dx Solution Although we know how to differentiate sec(x)2 , it isn’t likely to simplify the problem. The integral of sec(x)2 is tan(x), so it seems that the following definitions v(x) = x and du = sec(x)2 dx will help us solve the problem. We find dv =1 dx and u(x) = tan(x) This yields Z Z 2 x sec(x) dx = x tan(x) − 1 · tan(x)dx In order to evaluate the second integral we rewrite tan(x) = Define u = cos(x) sin(x) cos(x) and use substitution. which implies du = − sin(x)dx Now we can evaluate Z Z sin(x) 1 dx = − du = − ln(|u|) + c = − ln(| cos(x)|) + c cos(x) u Substituting into the original expression we find Z x sec(x)2 dx = x tan(x) − (− ln(| cos(x)|) + c) = x tan(x) + ln(| cos(x)|) + c Z Example 4 Evaluate x2 sin(x)dx Solution Here we want to differentiate x2 and integrate sin(x). If we were to integrate x2 , it would only yield higher powers of x, which only complicate the problem. However, if we differentiate x2 , then we can try and make it vanish (even if we need to integrate by parts twice) and we’ll be left with something less complicated to integrate, as the sin(x) will not become more complicated by integration. v(x) = x2 and This leads to du = sin(x) dx dv = 2x dx and u(x) = − cos(x) Using integration by parts Z Z Z 2 2 2 x sin(x)dx = x · (− cos(x)) − (− cos(x)) · 2xdx = −x cos(x) + 2 x cos(x)dx In order to evaluate the integral on the right we’ll need to use integration by parts once again setting v(x) = x and du = cos(x) dx which implies dv =1 dx and u(x) = sin(x) Now we can fine Z Z x cos(x)dx = x sin(x) − 1 · sin(x)dx = x sin(x) − (− cos(x)) + c = x sin(x) + cos(x) + c Substituting back into the original expression Z x2 sin(x)dx = −x2 cos(x) + 2(x sin(x) + cos(x) + c) = (2 − x2 ) cos(x) + 2x sin(x) + c