Prepared by Professor Antonino De Martino (Polytechnic University of Milan) and Dr. Luana Manfredini.

Basic Integral Formulas

Indefinite Integrals

\(\int k \, dx = kx + c\) \(\int x^\alpha \, dx = \frac{x^{\alpha +1}}{\alpha +1} + c\) \(\int \frac{1}{x} \, dx = \log |x| + c\) \(\int a^{kx} \, dx = \frac{a^{kx}}{k \log a} + c\) \(\int (f(x))^\alpha f'(x) \, dx = \frac{(f(x))^{\alpha +1}}{\alpha +1} + c\) \(\int \frac{f'(x)}{f(x)} \, dx = \log|f(x)| + c\) \(\int a^{f(x)} f'(x) \, dx = \frac{a^{f(x)}}{\log a} + c\)

Trigonometric and Hyperbolic Functions

\[\int \cos x \, dx = \sin x + c\] \[\int \sin x \, dx = -\cos x + c\] \[\int f'(x) \cos f(x) \, dx = \sin f(x) + c\] \[\int f'(x) \sin f(x) \, dx = -\cos f(x) + c\] \[\int \frac{1}{\cos^2 x} \, dx = \tan x + c\] \[\int \frac{1}{\sin^2 x} \, dx = -\cot x + c\] \[\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + c\] \[\int \frac{1}{1 + x^2} \, dx = \arctan x + c\] \[\int \frac{f'(x)}{1 + f(x)^2} \, dx = \arctan f(x) + c\] \[\int \cosh x \, dx = \sinh x + c\] \[\int \sinh x \, dx = \cosh x + c\]

Useful Trig Identities

\(\cos^2 x = \frac{1 + \cos(2x)}{2}\) \(\sin^2 x = \frac{1 - \cos(2x)}{2}\)

Integration Techniques

Integration by parts:

\[\int f(x) g'(x) \, dx = f(x)g(x) - \int f'(x) g(x) \, dx\]

Substitution rule:

\[\int f(x) \, dx = \int f(g(t)) g'(t) \, dt \quad \text{with } x = g(t)\]

Notable Improper Integrals

  1. If ( \alpha > 0 ):
\[\int_0^\alpha \frac{1}{x^p} \, dx = \begin{cases} \text{converges if } p < 1 \\\\ \text{diverges if } p \geq 1 \end{cases}\]
  1. If ( \alpha > 0 ):
\[\int_\alpha^{+\infty} \frac{1}{x^p} \, dx = \begin{cases} \text{converges if } p > 1 \\\\ \text{diverges if } p \leq 1 \end{cases}\]
  1. If ( \alpha > 1 ):
\[\int_1^\alpha \frac{1}{(\log x)^p} \, dx = \begin{cases} \text{converges if } p < 1 \\\\ \text{diverges if } p \geq 1 \end{cases}\]

Exercises

Exercise 1

Evaluate the integral:

\[\int_{\frac{1}{e}}^e \frac{\log(|\log x|)}{x} \, dx\]

Exercise 2

Evaluate the integral:

\[\int \frac{1}{\sin^2 x \cos^2 x} \, dx\]

Exercise 3

Evaluate the definite integral:

\[\int_{-1}^{1} \sqrt{1 - x^2} \, dx\]

Exercise 4

Evaluate the definite integral:

\[\int_{-1}^{1} \frac{1}{\sqrt{4 - x^2}} \, dx\]

Exercise 5

Evaluate the indefinite integral:

\[\int \frac{x^2}{x^2 + 2} \, dx\]

Exercise 6

Evaluate the indefinite integral:

\[\int \frac{1}{\sin x} \, dx\]

Exercise 7

Evaluate the definite integral:

\[\int_{-2}^{2} \sqrt{4 - x^2} \, dx\]

Exercise 8

Evaluate the indefinite integral:

\[\int \sin x \, e^x \, dx\]

Exercise 9

Evaluate the definite integral:

\[\int_0^1 e^{2x} \log(1 + e^x) \, dx\]

Exercise 10

Evaluate the definite integral:

\(\int_1^e \sin(\log x) \, dx\)

Solutions

Exercise 1

Evaluate the integral:

\[\int_{\frac{1}{e}}^e \frac{\log(|\log x|)}{x} \, dx\]

Solution:

We split the integral at the point where the argument of the logarithm changes sign:

\[\int_{\frac{1}{e}}^e \frac{\log(|\log x|)}{x} \, dx = \int_{\frac{1}{e}}^1 \frac{\log(-\log x)}{x} \, dx + \int_1^e \frac{\log(\log x)}{x} \, dx\]

In the first part, let

\[t = -\log x, \quad dt = -\frac{dx}{x}\]

Then:

\[\int_{\frac{1}{e}}^1 \frac{\log(-\log x)}{x} \, dx = - \int_1^0 \log t \, dt = \int_0^1 \log t \, dt\]

For the second part, let

\[t = \log x, \quad dt = \frac{dx}{x}\]

So:

\[\int_1^e \frac{\log(\log x)}{x} \, dx = \int_0^1 \log t \, dt\]

Therefore:

\[\int_{\frac{1}{e}}^e \frac{\log(|\log x|)}{x} \, dx = 2 \int_0^1 \log t \, dt\]

Using integration by parts:

\[\int_0^1 \log t \, dt = \lim_{a \to 0^+} \left( t \log t - t \right)\bigg|_a^1 = \lim_{a \to 0^+} \left( -1 - a \log a + a \right) = -1\]

Thus:

\[\int_{\frac{1}{e}}^e \frac{\log(|\log x|)}{x} \, dx = -2\]

Exercise 2

Evaluate the integral:

\[\int \frac{1}{\sin^2 x \cos^2 x} \, dx\]

Solution:

We use the identity:

\[\sin^2 x + \cos^2 x = 1\]

Then:

\[\frac{1}{\sin^2 x \cos^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}\]

So:

\[\int \frac{1}{\sin^2 x \cos^2 x} \, dx = \int \frac{1}{\cos^2 x} \, dx + \int \frac{1}{\sin^2 x} \, dx = \tan x - \cot x + c\]

Exercise 3

Evaluate the definite integral:

\[\int_{-1}^{1} \sqrt{1 - x^2} \, dx\]

Solution:

Let

\[x = \sin t, \quad dx = \cos t \, dt\]

Then:

\[\int_{-1}^{1} \sqrt{1 - x^2} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 t \, dt\]

Use the identity:

\[\cos^2 t = \frac{1 + \cos(2t)}{2}\]

So:

\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2t)}{2} \, dt = \left( \frac{t}{2} + \frac{\sin(2t)}{4} \right)\bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2}\]

Exercise 4

Evaluate the definite integral:

\[\int_{-1}^{1} \frac{1}{\sqrt{4 - x^2}} \, dx\]

Solution:

We write:

\[\int_{-1}^{1} \frac{1}{\sqrt{4 - x^2}} \, dx = \int_{-1}^{1} \frac{1}{2 \sqrt{1 - \left( \frac{x}{2} \right)^2}} \, dx\]

Let

\[t = \frac{x}{2}\]

Then:

\[\int_{-1}^{1} \frac{1}{\sqrt{4 - x^2}} \, dx = 2 \int_0^{\frac{1}{2}} \frac{1}{\sqrt{1 - t^2}} \, dt = 2 \arcsin t \bigg|_0^{\frac{1}{2}} = \frac{\pi}{3}\]

Exercise 5

Evaluate the indefinite integral:

\[\int \frac{x^2}{x^2 + 2} \, dx\]

Solution:

We rewrite the numerator:

\[\int \frac{x^2 + 2 - 2}{x^2 + 2} \, dx = \int 1 \, dx - \int \frac{2}{x^2 + 2} \, dx\]

Now:

\[\int \frac{2}{x^2 + 2} \, dx = \int \frac{1}{\left( \frac{x}{\sqrt{2}} \right)^2 + 1} \, dx\]

Let

\[t = \frac{x}{\sqrt{2}}, \quad dt = \frac{dx}{\sqrt{2}}\]

Then:

\[\sqrt{2} \int \frac{1}{1 + t^2} \, dt = \sqrt{2} \arctan t + c\]

So the result is:

\[x - \sqrt{2} \arctan\left( \frac{x}{\sqrt{2}} \right) + c\]

Exercise 6

Evaluate the indefinite integral:

\[\int \frac{1}{\sin x} \, dx\]

Solution:

Using the identity:

\[\int \frac{1}{\sin x} \, dx = \log \left| \tan \frac{x}{2} \right| + c\]

Exercise 7

Evaluate the definite integral:

\[\int_{-2}^{2} \sqrt{4 - x^2} \, dx\]

Solution:

By symmetry:

\[\int_{-2}^{2} \sqrt{4 - x^2} \, dx = 2 \int_0^2 \sqrt{4 - x^2} \, dx\]

Let

\[x = 2 \sin t, \quad dx = 2 \cos t \, dt\]

Then:

\[8 \int_0^{\frac{\pi}{2}} \cos^2 t \, dt = 8 \left( \frac{t}{2} + \frac{\sin(2t)}{4} \right) \bigg|_0^{\frac{\pi}{2}} = 2\pi\]

Exercise 8

Evaluate the indefinite integral:

\[\int \sin x \, e^x \, dx\]

Solution:

Integration by parts twice:

Let

\[I = \int \sin x \, e^x \, dx\]

Then:

\[I = \sin x \, e^x - \int e^x \cos x \, dx\]

and

\[\int e^x \cos x \, dx = e^x \cos x + I\]

So:

\[2I = e^x (\sin x - \cos x)\]

and therefore:

\[\int \sin x \, e^x \, dx = \frac{e^x (\sin x - \cos x)}{2} + c\]

Exercise 9

Evaluate the definite integral:

\[\int_0^1 e^{2x} \log(1 + e^x) \, dx\]

Solution:

Let

\[t = e^x, \quad dx = \frac{dt}{t}\]

Bounds: when ( x = 0 ), ( t = 1 ); when ( x = 1 ), ( t = e )

The integral becomes:

\[\int_1^e t \log(1 + t) \, dt\]

Use integration by parts:

\[\int_1^e t \log(1 + t) \, dt = \frac{t^2}{2} \log(1 + t) \bigg|_1^e - \frac{1}{2} \int_1^e \frac{t^2}{1 + t} \, dt\]

Divide the integrand:

\[\frac{t^2}{1 + t} = t - 1 + \frac{1}{1 + t}\]

Then:

\[\int_1^e \frac{t^2}{1 + t} \, dt = \left( \frac{t^2}{2} - t + \log(1 + t) \right) \bigg|_1^e\]

Putting it all together:

\[\int_0^1 e^{2x} \log(1 + e^x) \, dx = \frac{e^2}{2} \log(1 + e) - \frac{e^2}{4} + \frac{e}{2} - \frac{1}{4} - \frac{\log(1 + e)}{2}\]

Exercise 10

Evaluate the definite integral:

\[\int_1^e \sin(\log x) \, dx\]

Solution:

We integrate by parts:

Let:

\[u = \sin(\log x), \quad dv = dx\]

Then:

\[\int_1^e \sin(\log x) \, dx = x \sin(\log x) \Big|_1^e - \int_1^e x \cdot \frac{d}{dx}[\sin(\log x)] \cdot \frac{1}{x} \, dx\]

We compute:

\[\frac{d}{dx}[\sin(\log x)] = \frac{1}{x} \cos(\log x)\]

So the integral becomes:

\[e \sin 1 - \int_1^e \cos(\log x) \, dx\]

Now we integrate by parts again:

Let:

\[\int_1^e \cos(\log x) \, dx = x \cos(\log x) \Big|_1^e - \int_1^e \frac{1}{x} \cdot (-\sin(\log x)) \cdot x \, dx = e \cos 1 - \int_1^e \sin(\log x) \, dx\]

Putting all together:

\[\int_1^e \sin(\log x) \, dx = e \sin 1 - \left( e \cos 1 - \int_1^e \sin(\log x) \, dx \right)\]

Simplifying:

\[2 \int_1^e \sin(\log x) \, dx = e(\sin 1 - \cos 1) + 1\]

Therefore:

\[\int_1^e \sin(\log x) \, dx = \frac{e(\sin 1 - \cos 1)}{2} + \frac{1}{2}\]