Worked examples · Sheet 3

Integration by substitution

Theory used: Integration by substitution

Exercise 1

Compute

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

With the substitution $t = x$ we have $d t = \frac{1}{2 x} d x$ , that is $\frac{d x}{x} = 2 d t$ , therefore

\int \frac{e ^{x}}{x} d x = 2 \int e^{t} d t = 2 e^{t} + c = 2 e^{x} + c .

The integrand is a function of $x$ accompanied by the factor $\frac{1}{x}$ , which up to the constant $2$ is the derivative of $x$ . We therefore set $t = x$ and express the differential: from $d t = \frac{d x}{2 x}$ it follows that $\frac{d x}{x}$ is replaced by $2 d t$ . The integral in the variable $t$ is immediate and at the end one returns to the variable $x$ with $t = x$ . Differentiating $2 e^{x}$ gives $2 e^{x} \cdot \frac{1}{2 x} = \frac{e ^{x}}{x}$ , therefore the result is correct.

Exercise 2

Compute

\int \frac{x}{x + 1} d x .

With the substitution $t = x + 1$ we have $x = t^{2} - 1$ and $d x = 2 t d t$ , therefore

\int \frac{x}{x + 1} d x = \int \frac{t ^{2} - 1}{t} \cdot 2 t d t = 2 \int (t^{2} - 1) d t = \frac{2}{3} t^{3} - 2 t + c = \frac{2}{3} (x + 1) x + 1 - 2 x + 1 + c

The root is eliminated with the substitution $t = x + 1$ , also written in the inverse form $x = t^{2} - 1$ , from which $d x = 2 t d t$ . After the substitution the factor $t$ in the denominator cancels with the factor $2 t$ of the differential, therefore what remains is the integral of the polynomial $2 (t^{2} - 1)$ , which is computed with the table. To return to the variable $x$ one substitutes $t = x + 1$ , and $t^{3}$ becomes $(x + 1) x + 1$ . Differentiating the result gives $x + 1 - \frac{1}{x + 1} = \frac{( x + 1 ) - 1}{x + 1} = \frac{x}{x + 1}$ , which is the integrand.

Exercise 3

Compute

\int 1 - x^{2} d x .

With the inverse substitution $x = sin t$ , $t \in [- \frac{π}{2}, \frac{π}{2}]$ , we have $d x = cos t d t$ and $1 - sin^{2} t = cos t$ , therefore

\int 1 - x^{2} d x = \int cos^{2} t d t = \int \frac{1 + cos 2 t}{2} d t = \frac{t}{2} + \frac{sin 2 t}{4} + c = \frac{t + sin t cos t}{2} + c .

Returning to the variable $x$ with $t = arcsin x$ , $sin t = x$ and $cos t = 1 - x^{2}$ :

\int 1 - x^{2} d x = \frac{arcsin x + x 1 - x ^{2}}{2} + c .

The root $1 - x^{2}$ suggests the inverse substitution $x = sin t$ , because $1 - sin^{2} t = cos^{2} t$ . On the interval chosen for $t$ the cosine is nonnegative, therefore $cos^{2} t = cos t$ without absolute value. The integral of $cos^{2} t$ is computed with the half-angle formula $cos^{2} t = \frac{1 + cos 2 t}{2}$ , and the integral of $cos 2 t$ is $\frac{sin 2 t}{2}$ because the derivative of $sin 2 t$ is $2 cos 2 t$ . With the double-angle formula $sin 2 t = 2 sin t cos t$ the result is rewritten in terms of $sin t$ and $cos t$ . The substitution is invertible on the chosen interval, therefore $t = arcsin x$ ; since $sin t = x$ , we have $cos t = 1 - x^{2}$ and the result returns to the variable $x$ .