Worked examples · Sheet 2

Integration by parts

Exercise 1

Compute

\int x e^{x} d x .

\int x e^{x} d x = x e^{x} - \int 1 \cdot e^{x} d x = x e^{x} - e^{x} + c = (x - 1) e^{x} + c

The integrand is the product of a polynomial and an exponential, therefore one integrates by parts differentiating the polynomial: we set $f (x) = x$ and $g^{'} (x) = e^{x}$ , from which $f^{'} (x) = 1$ and $g (x) = e^{x}$ . The new integral is that of $e^{x}$ , which is immediate. Differentiating $(x - 1) e^{x}$ gives $e^{x} + (x - 1) e^{x} = x e^{x}$ , therefore the result is correct.

Exercise 2

Compute

\int x ln x d x .

\int x ln x d x = \frac{x ^{2}}{2} ln x - \int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x = \frac{x ^{2}}{2} ln x - \frac{1}{2} \int x d x = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + c

When the logarithm appears it is convenient to differentiate it, because its derivative $\frac{1}{x}$ is a rational function. We therefore set $f (x) = ln x$ and $g^{'} (x) = x$ , from which $f^{'} (x) = \frac{1}{x}$ and $g (x) = \frac{x ^{2}}{2}$ . In the new integral the product $\frac{x ^{2}}{2} \cdot \frac{1}{x}$ simplifies to $\frac{x}{2}$ , hence what remains is the immediate integral of $x$ multiplied by $\frac{1}{2}$ , which equals $\frac{x ^{2}}{4}$ .

Exercise 3

Compute

\int e^{x} sin x d x .

Setting $I = \int e^{x} sin x d x$ and integrating by parts twice, we obtain

I = e^{x} sin x - \int e^{x} cos x d x = e^{x} sin x - (e^{x} cos x + \int e^{x} sin x d x) = e^{x} (sin x - cos x) - I .

From the equality $I = e^{x} (sin x - cos x) - I$ it follows that $2 I = e^{x} (sin x - cos x)$ , therefore

\int e^{x} sin x d x = \frac{e ^{x} ( sin x - cos x )}{2} + c .

In the first integration by parts we set $f (x) = sin x$ and $g^{'} (x) = e^{x}$ , therefore the integral of $e^{x} cos x$ appears. In this one we set $f (x) = cos x$ and $g^{'} (x) = e^{x}$ ; since the derivative of $cos x$ is $- sin x$ , in the integral of $e^{x} cos x$ the original integral $I$ reappears, added to $e^{x} cos x$ . The equality obtained is an equation in the unknown $I$ : moving $I$ to the left-hand side gives $2 I = e^{x} (sin x - cos x)$ and one divides by $2$ . The constant $c$ is added at the end, because the result represents the set of all antiderivatives.