Reminders

Primitive of \(F\): \(F’ = f\)
Every continuous function on \(I\) has at least one primitive on \(I\).
2 primitives of the same function differ by a constant.
Linearity:
$$\int (f(x)+g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx$$
$$\int \lambda.f(x) \, dx = \lambda.\int f(x) \, dx$$
The integral between \(a\) and \(b\) of \(f\) is:
$$\int_a^b f(x) \,dx = F(b)-F(a)$$

Chasles’ relationship

$$\int_a^c f(x) \,dx = \int_a^b f(x) \,dx + \int_b^c f(x) \,dx $$
$$\int_a^a f(x) \,dx =0$$
$$\int_b^a f(x) \,dx = – \int_a^b f(x) \,dx$$

Positivity

$$ \forall x \in [a;b], f(x)\geq 0 \Rightarrow \int_a^b f(x) \,dx \geq 0 $$

Integration by parts

Let \(u\) and \(v\) be 2 differentiable functions on \([a;b]\) and continuous differentiable on this same interval:
$$\int_a^b u(x).v'(x) \,dx = [u(x).v(x)]_a^b – \int_a^b u'(x).v(x) \, dx$$
Or:
$$\int_a^b u'(x).v(x) \,dx = [u(x).v(x)]_a^b – \int_a^b u(x).v'(x) \, dx$$

Change of variable

A differentiable function \(φ\) on \(I\) and \(a,b ∈ I\). Let \(J\) be an interval containing \(φ(I)\).
Let \(f\) be defined and continuous on \(J\), and \(F\) be a primitive of \(f\) on \(J\):
$$\int_a^b f(φ(x)).φ'(x) \,dx = \int_{φ(a)}^{φ(b)} f(u) \,du $$
To remember :

• Change bounds
• Change of variable in one go
• Modify the \(dx\):

If \(x = φ(t)\) then \(dx = φ'(x)dt\)
If \(u = φ(x)\) then \(du = φ'(x)dx\)
\(dx\) cannot be replaced by \(\frac{du}φ'(x)\) because \(u\) and \(x\) do not coexist in the same expression.
Example:
$$ I=\int_{\frac13}^{\frac12} x\sqrt{1-x^2} \,du $$
Let \(x^2=u\), so \(2xdx=du\) and \(xdx=\frac{du}2\)
For \(x=\frac{1}3\), \(u=\frac{1}9\) and for \(x=\frac{1}2\), \(u=\frac{1} 4\)
So
$$ I=\int_{\frac13}^{\frac12} \sqrt{1-x^2} \,(xdx) = \int_{\frac14}^{\frac19} \sqrt{1-u} \, \frac{du}2 $$

Change of variable in a primitive calculation

Remember:

• When we make a change of variable, we don’t care about the bounds
• Remember to return to the initial variable at the end
• Think about the constant
• Don’t forget \(dx\) (or \(dt\), or \(du\) …)

Example: Determine the primitives of \(f:x \rightarrow \sqrt{1+x^2}\) on \(\mathbb{R}\)
\(x=sinh⁡(t)\) because \(cosh^2⁡(t)-sinh^2(t)=1\), so \(1+sinh^2⁡(t)=cosh^2( t)\)
$$ F(x)=\int f(x) \,dx = \int \sqrt{1+sinh^2⁡(t)}*(cosh(t) \,dt)= \int \sqrt{cosh^ 2⁡(t)}*(cosh(t) \,dt)= \int cosh^2(t) \,dt $$
We know that \(cosh⁡(t)≥1>0\) for all \(t∈\mathbb{R}\)
$$ F(x) = \int cosh^2(t) \,dt = \int (\frac{e^{t}+e^{-t}}2)^2 \,dt = \int \frac {e^{2t}+e^{-2t}+2}4 \,dt$$
$$F(x)=\int (\frac{1}2 + \frac{cosh(2t)}2) \,dt = \frac{t}2 + \frac{sinh(2t)}4 +C= \frac{t}2 + \frac{sinh(t)*cosh(t)}2 +C$$
Because \(sinh⁡(2t)=2 sinh⁡(t)*cosh⁡(t)\)
Returning to the variable \(x\) knowing that: \(x=sinh⁡(t)\), we have: \(t=arg⁡sinh⁡(x)\), therefore:
$$ F(x)=\frac{arg⁡sinh⁡(x)}2 + \frac{sinh⁡(arg⁡sinh⁡(x))*cosh⁡(arg⁡sinh⁡(x))}2 + C = \frac{arg⁡sinh⁡(x)}2 + \frac{x\sqrt{1+x^2}}2 + C$$ (\(cosh⁡(x) = arg⁡sinh⁡(x) =\sqrt{1+x^2}\))
It has been proven that:
$$ \int \sqrt{1+x^2} \,dt = \frac{1}2*(arg⁡sinh⁡(x)+x\sqrt{1+x^2})+ C$$

Integration of an even or odd function on a “symmetric” interval

\(f\) pair \(\rightarrow \int_{-a}^a f(x) \,dx =2\int_0^a f(x) \,dx\)
\(f\) odd \(\rightarrow \int_{-a}^a f(x) \,dx =0\)

Usual primitives

f(x)	F(x)	Definition interval	Conditions
\(x^n\)	\(\frac{x^{n+1}}{n+1}\)	\(\mathbb{R}\)	\(\text{if } n \in \mathbb{R}\)
\(x^n\)	\(\frac{x^{n+1}}{n+1}\)	\(]-\infty;0[\text{ and }]0;+\infty[\)	\(\text{if } n \in \mathbb{Z} \setminus \{1 \} \)
\(\frac{x^{n+1}}{n+1} = x^{-1}\)	\(ln\vert x \vert\)	\(]-\infty;0[\text{ and }]0;+\infty[\)
\(x^{\alpha}\)	\(\frac{x^{\alpha +1}}{\alpha +1}\)	\(]0;+\infty[\)	\(\text{if } \alpha \in \mathbb{R} \setminus \{1 \}\)
\(\frac{1}{x^2+1}\)	\(arctan⁡(x)\)	\(\mathbb{R}\)
\(x^n\)	\(arcsin(x)\)	\(]-1;1[\)
\(\frac{1}{x^2+1}\)	\(arctan⁡(x)\)	\(\mathbb{R}\)
\(e^x\)	\(e^x\)	\(\mathbb{R}\)
\(sin(x)\)	\(-cos(x)\)	\(\mathbb{R}\)
\(cos(x)\)	\(sin(x)\)	\(\mathbb{R}\)
\(sinh(x)\)	\(cosh(x)\)	\(\mathbb{R}\)
\(cosh(x)\)	\(sinh(x)\)	\(\mathbb{R}\)
\(\frac{1}{\sqrt{x^2+1}}\)	\(argsinh(x)\)	\(\mathbb{R}\)
\(\frac{1}{\sqrt{x^2-1}}\)	\(argcosh(x)\)	\(]1;+\infty[\)
\(g'(ax+b)\)	\(\frac{g(ax+b)}{a}\)		\(a \neq 0\)

Tagged: orci, lectus, varius, turpis