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Mathematics · Ch 17 — Definite Integrals

Integration by Parts

17.6

Integration by Parts

Integration by Parts

Integration by parts integrates products of functions. It is derived directly from the product rule of differentiation and transforms a difficult integral into a simpler one.

The Fundamental Formula

Let uu and vv be two differentiable functions of xx. From the product rule:

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Integrating both sides with respect to xx:

uv=∫udvdx dx+∫vdudx dxuv = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx

Rearranging gives the integration by parts formula:

∫udvdx dx=uv−∫vdudx dx\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx

Standard Form

Let u=f(x)u = f(x) and dvdx=g(x)\frac{dv}{dx} = g(x). Then dudx=f′(x)\frac{du}{dx} = f'(x) and v=∫g(x) dxv = \int g(x)\,dx. Substituting:

Integration by Parts Formula

∫f(x)g(x) dx=f(x)∫g(x) dx−∫[f′(x)∫g(x) dx]dx\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x)\int g(x)\,dx\right]dx

In words: the integral of the product equals (first function) × (integral of the second function) minus the integral of [(derivative of the first function) × (integral of the second function)].

Choosing the First and Second Functions

The choice of first function (ff) and second function (gg) is crucial — a wrong choice can make the integral more complicated instead of simpler.

Tip

Guidelines for choosing the first function:

  • If one function is a power of xx or a polynomial in xx, take it as the first function.
  • If one function is an inverse trigonometric function or a logarithmic function, take that as the first function.

Important Remarks

Remark (i): Applicability

Integration by parts is not applicable to all products of functions. For example, ∫xsin⁡x dx\int\sqrt{x}\sin x\,dx cannot be evaluated by this method because there is no function whose derivative is xsin⁡x\sqrt{x}\sin x.

Remark (ii): The Constant of Integration

When finding the integral of the second function, we do not add a constant of integration — it cancels out in the final result.

Verification: Suppose we write ∫cos⁡x dx=sin⁡x+k\int\cos x\,dx = \sin x + k (where kk is any constant). Then:

∫xcos⁡x dx=x(sin⁡x+k)−∫1⋅(sin⁡x+k) dx\int x\cos x\,dx = x(\sin x + k) - \int 1\cdot(\sin x + k)\,dx

=xsin⁡x+kx−∫sin⁡x dx−∫k dx= x\sin x + kx - \int\sin x\,dx - \int k\,dx

=xsin⁡x+kx−(−cos⁡x)−kx+C=xsin⁡x+cos⁡x+C= x\sin x + kx - (-\cos x) - kx + C = x\sin x + \cos x + C

The kxkx terms cancel, confirming that adding a constant is unnecessary.

Standard Results

IntegralResult
∫xcos⁡x dx\int x\cos x\,dxxsin⁡x+cos⁡x+Cx\sin x + \cos x + C
∫log⁡x dx\int\log x\,dxxlog⁡x−x+Cx\log x - x + C
∫xex dx\int xe^x\,dxex(x−1)+Ce^x(x-1) + C