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Mathematics · Ch 1 — Integrals

Integration by Parts

1.6

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