Integration by Parts

By the end of this lesson, you will be able to apply the Integration by Parts formula, use the LIATE rule to choose $u$ and $dv$, handle repeated applications and circular integrals.

The Formula

Integration by Parts reverses the Product Rule:

$$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$

Integrate both sides:

$$\int u,dv = uv - \int v,du$$

The LIATE Rule for Choosing $u$

Choose $u$ to be the type that appears earlier in the list:

L — Logarithms ($\ln x$, $\log x$)

I — Inverse trig ($\sin^{-1}x$, $\tan^{-1}x$)

A — Algebraic ($x^n$, polynomials)

T — Trigonometric ($\sin x$, $\cos x$)

E — Exponential ($e^x$, $a^x$)

The other factor becomes $dv$.

Example 1 — Basic (Algebraic × Exponential)

$$\int xe^x,dx$$

$u = x$, $dv = e^x,dx$; $du = dx$, $v = e^x$

$= xe^x - \int e^x,dx = xe^x - e^x + C = e^x(x-1) + C$

Example 2 — Algebraic × Trigonometric

$$\int x\cos x,dx$$

$u = x$, $dv = \cos x,dx$; $du = dx$, $v = \sin x$

$= x\sin x - \int\sin x,dx = x\sin x + \cos x + C$

Example 3 — Logarithm

$$\int\ln x,dx$$

$u = \ln x$, $dv = dx$; $du = dx/x$, $v = x$

$= x\ln x - \int x \cdot \frac{dx}{x} = x\ln x - x + C = x(\ln x - 1) + C$

Example 4 — Repeated Integration by Parts

$$\int x^2 e^x,dx$$

Apply twice: first $u = x^2$, then $u = 2x$.

Round 1: $x^2 e^x - 2\int xe^x,dx$

From Example 1: $\int xe^x,dx = e^x(x-1)$

Result: $x^2 e^x - 2e^x(x-1) + C = e^x(x^2-2x+2) + C$

Example 5 — Circular Integration (Solving Algebraically)

$$\int e^x\cos x,dx$$

Apply parts: $u = e^x$, $dv = \cos x,dx$

$= e^x\sin x - \int e^x\sin x,dx$

Apply parts again to $\int e^x\sin x,dx$: $u = e^x$, $dv = \sin x,dx$

$= e^x\sin x - \left(-e^x\cos x + \int e^x\cos x,dx\right)$

$= e^x\sin x + e^x\cos x - \int e^x\cos x,dx$

Let $I = \int e^x\cos x,dx$:

$I = e^x\sin x + e^x\cos x - I$

$2I = e^x(\sin x + \cos x)$