Standard antiderivatives and general integration methods

Since each antiderivative of a given function is equal up to a constant, we will use the unified notation ${\displaystyle \int^x} f(t) \ dt $,meaning this family of primitives (or general antiderivative).

We will sometimes use the capital letter of a function to signify its antiderivative.

Recall on standard antiderivatives

A certain amount of antiderivative can be directly calculated by taking the reverse path of the derivative or by an integration by parts.

For others, each case must be considered as a specific case.

General integration methods

Integration by parts

$$ \forall (a,b) \in D_f^2,$$

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b - \int_{a}^b (fg') \hspace{0.2em}dt $$

Integration by substitution

Let $\phi : t \longmapsto \phi(t) $ be a function of class $\mathcal{C}^1$ on an interval $ J \subset f(I) $.

The integration by substitution is the transposition of the derivative of a composite function, but applied to integral calculus.

$$ \forall (a,b) \in D_f^2, \ $$

$$ \int_{a}^b \hspace{0.2em} \Bigl(f \circ \phi(t)\Bigr) \hspace{0.2em} \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(u) \hspace{0.2em}du $$

$$ with \enspace \Biggl \{ \begin{align*} u = \phi(t) \\ du = \phi '(t) \hspace{0.2em} dt \end{align*} $$

General integration methods recap table

Demonstrations

Recall on standard antiderivatives

This table resumes antiderivatives directly determined from the inverse operation of the derivative, namey:

$$ f(x) = \int^x F'(t)dt \ \Longleftrightarrow \ F(x) = \int^x f(t)dt $$

$$ \underline{condition} $$	$$ \underline{general \ antiderivative} $$
$$ \underline{standard \ functions} $$
$$ \forall x \in \mathbb{R}, $$	$$ \int^x dt = x$$
$$ \forall x \in \mathbb{R}, $$	$$ \int^x t \ dt = \frac{x^2}{2}$$
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*_+,$$	$$ \int^x \sqrt{t} \ dt = \frac{2}{3} x^{\frac{3}{2}}$$
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*_+,$$	$$ \int^x \frac{1}{\sqrt{t}} \ dt = 2\sqrt{x} $$
$$ when \ x \ is \ defined $$	$$ \int^x t^n \ dt = \frac{x^{n+1}}{n+1}$$
$$\forall x \in \mathbb{R}, \enspace \forall n \in \mathbb{R_+^*},$$	$$ \int^x n^t \ dt = \frac{ n^x}{ln(n)} $$
$$ \forall x \in \mathbb{R}, $$	$$ \int^x e^t \ dt = e^x$$
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, $$	$$ \int^x\frac{dt}{t} = ln\|x\| $$
$$ \forall x \in \mathbb{R^*}, \enspace \forall n \in \mathbb{R^+}, $$	$$ \int^x log_n\|t\| \ dt = ln(n) \times log_n\|x\| $$
$\Longrightarrow \ $all derivatives of standard functions	$$ $$
$$ \underline{trigonometric \ functions} $$
$$ \forall x \in [-1, \hspace{0.2em} 1], $$	$$ \int^x \frac{1}{\sqrt{1 - t^2}} \ dt = arcsin(x) = - arccos(x) $$
$$ \forall x \in \mathbb{R}, $$	$$ \int^x \frac{1}{\sqrt{1 + t^2}} \ dt = arcsinh(x) $$
$$ \forall x \in [1, \hspace{0.1em} +\infty[, $$	$$ \int^x \frac{1}{\sqrt{t^2 - 1}} \ dt = arccosh(x) $$
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr] $$	$$ \int^x \Bigl[ 1 + tan^2(t) \Bigr] \ dt = \int^x sec^2(t) \ dt = tan(x) $$
$$ \forall x \in \mathbb{R}, $$	$$ \int^x \frac{1}{1 + t^2} \ dt = arctan(x) $$
$$ \forall x \in \mathbb{R}, $$	$$ \int^x sech^2(t) \ dt = tanh(x) $$
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$	$$ \int^x \frac{1}{1 - t^2} \ dt = arctanh(x) $$
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$	$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ t^2}}} \Biggl] \ dt = -arccosec(x) $$
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \left \{ 0 \right \} \Bigr] , $$	$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{1 + \frac{1}{ t^2}}} \Biggl] \ dt = -arccosech(x) $$
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$	$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ t^2}}} \Biggl] \ dt = arcsec(x) $$
$$ \forall x \in \hspace{0.1em} ]0, \hspace{0.1em} 1], $$	$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{\frac{1}{ t^2}- 1}} \Biggl] \ dt = -arcsech(x) $$
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], $$	$$ \int^x cosec^2(t) \ dt = -cotan(x) $$
$$ \forall x \in \mathbb{R}, $$	$$ \int^x \frac{1}{1+t^2} \ dt = -arccotan(x) $$
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, $$	$$ \int^x \Bigl[ 1 - cotan^2(t) \Bigr] \ dt = \int^x \Bigl[ -cosech^2(t) \Bigr] \ dt = cotanh(x) $$
$$ \forall \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , $$	$$ \int^x \frac{1}{1 - t^2} \ dt = arcccotanh(x) $$
$\Longrightarrow \ $all trigonometric derivatives	$$ $$
$\Longrightarrow \ $all trigonometric antiderivatives	$$ $$
$$ \underline{operations \ on \ functions} $$
$$ \forall a \in \mathbb{R}, \ \forall x \in \hspace{0.05em} \mathcal{D}_f, $$	$$ \int^x f(at) \ dt = \frac{1}{a} F(ax) $$
$$ \forall (\lambda, \mu)) \in \hspace{0.05em} \mathbb{R}^2, \ \forall \bigl(u(x), v(x)\bigr) \in \hspace{0.05em} \mathbb{R}^2, $$	$$ \int^x \biggl[ \lambda u(t) + \mu v(t) \biggr] \ dt = \lambda U(x) + \mu V(x) $$
$$ \forall u(x) \in \hspace{0.05em} \mathbb{R}^*, $$	$$ \int^x \frac{u'(t)}{u(t)} \ dt = ln\bigl\|u(x)\bigr\| $$
$$ \forall u(x) \in \hspace{0.05em} \mathbb{R}^*, $$	$$ \int^x \frac{u'(t)}{u^2(t)} \ dt = - \frac{1}{u(x)} $$
$$ \forall u(x) \in \hspace{0.05em} \mathbb{R}^*, $$	$$ \int^x \frac{u'(t)}{ 2\sqrt{u(t)}} \ dt = \sqrt{u(x)} $$
$\Longrightarrow \ $all derivatives of operations on functions	$$ $$

General integration methods

Integration by parts

With the derivative of a product, we do have:

$$ \left ( f g\right)' = f'g + g'f $$

$$ f'g = ( f g)' - g'f $$

Now, thanks to the property of linearity of the integral,

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \int_{a}^b (fg) -\int_{a}^b (fg') \hspace{0.2em}dt $$

And,

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b -\int_{a}^b (fg') \hspace{0.2em}dt $$

Then finally,

$$ \forall (a,b) \in I^2, \enspace a < b, $$

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b - \int_{a}^b (fg') \hspace{0.2em}dt $$

Integration by substitution

Let $f : x \longmapsto f(x)$ be a function of class $\mathcal{C}^1$ on an interval $I = [a,b]$.

As well, let $\phi : t \longmapsto \phi(t) $ be a function of class $\mathcal{C}^1$ on an interval $ J \subset f(I)$.

Condering the following integral $I(x)$:

$$ I(x) = \int_{a}^b f(x) \hspace{0.2em}dx = F(b) - F(a) \qquad(I(x)) $$

Performing the substitution $ x = \phi(t) $ in $ (I(x)) $, we do have now:

$$ x = \phi(t) \Longrightarrow \Biggl \{ \begin{align*} x \ \longrightarrow \ \phi(t) \\ dx \ \longrightarrow \ d \Bigl[ \phi(t)\Bigr] = \phi'(t) \ dt \end{align*} $$

Now considering $ \psi : x \longmapsto \psi(x) $, the reciprocal function of $ \phi$, we do have the following equivalence:

$$ x = \phi(t) \Longleftrightarrow t = \psi(x)$$

Then, when $x$ varies from $ a $ to $b$, $t$ varies from $ \psi(a) $ to $\psi(b)$.

So,

$$ I(t) = \int_{\psi(a)}^{\psi(b)} \Bigl(f \circ \phi(t)\Bigr) \ \phi'(t) \ dt \qquad(I(t)) $$

$$ I(t) = \Bigl[ f \circ \phi(t) \Bigr]_{\psi(a)}^{\psi(b)} $$

Both functions $\phi$ and $\psi$ being two reciprocal functions, they annihilate with each other.

$$ I(t) = F(b) - F(a) $$

Thus, both expressions $ (I(x)) $ and $ (I(t)) $ are equals.

$$ \int_{\psi(a)}^{\psi(b)} f[ \phi(t)\Bigr] \ \phi'(t) \ dt = \int_{a}^b f(x) \hspace{0.2em}dx \qquad (1) $$

In the end, transforming all bounds by performing the function $ \phi $, then $ (1)$ becomes $ (1')$:

$$ \int_{\phi(\psi(a))}^{\phi(\psi(b))} f[ \phi(t)\Bigr] \ \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(x) \hspace{0.2em}dx \qquad (1') $$

And,

$$ \int_{a}^{b} f[ \phi(t)\Bigr] \ \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(x) \hspace{0.2em}dx $$

For the sake of simplicity, we will use $u $ as a variable, and finally,

$$ \forall (a,b) \in D_f^2,$$

$$ \int_{a}^b \hspace{0.2em} \Bigl(f \circ \phi(t)\Bigr) \hspace{0.2em} \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(u) \hspace{0.2em}du $$

$$ with \enspace \Biggl \{ \begin{align*} u = \phi(t) \\ du = \phi '(t) \hspace{0.2em} dt \end{align*} $$

Be careful not to be confused with a direct variable change:

$$ \Biggl \{ \begin{align*} u = \phi(t) \\ du = \phi '(t) \hspace{0.2em} dt \end{align*} $$

in opposition to an indirect variable change:

$$ \Biggl \{ \begin{align*} t = \psi(u) \\ dt = \psi '(u) \hspace{0.2em} du \end{align*} $$

General integration methods recap table

Integration examples

Integration by parts examples

Example 1

$$ \alpha = \int_{0}^{\frac{\pi}{2}} t .sin(t) \hspace{0.2em}dt $$

We choose to take $f$ and $g'$ such as:

$$ \Biggl \{ \begin{align*} f(t) = t \\ g'(t) = dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = \frac{dt}{t} \\ g(t) = t.dt \end{align*} $$

$$ \alpha = \Bigl[-t.cos(t) \Bigr]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} - cos(t) \hspace{0.2em}dt $$

$$ \alpha = \Bigl[-t.cos(t) + sin(t) \Bigr]_{0}^{\frac{\pi}{2}} $$

$$ \alpha = - \frac{\pi}{2} \times 0 + 1 - (0\times 1 + 0 ) $$

$$ \alpha =1 $$

Example 2

$$ \beta = \int_{1}^{e} ln(t) \hspace{0.2em}dt $$

We choose to take $f$ and $g'$ such as:

$$ \Biggl \{ \begin{align*} f(t) = ln(t) \\ g'(t) = dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = \frac{dt}{t} \\ g(t) = t \end{align*} $$

$$ \beta = \Bigl[t.ln(t) \Bigr]_{1}^{e} - \int_{1}^{e} \hspace{0.2em}dt $$

$$ \beta = \Bigl[t.ln(t) \Bigr]_{1}^{e}- \Bigl[t \Bigr]_{1}^{e} $$

$$ \beta = e . ln(e) - ln(1) - (e -1) $$

$$ \beta =1 $$

Example 3: integrate a function of type $f(t) e^t$

Notably while solving ordinary differential equations, we may determine antiderivatives of type:

$$ \int^{x} f(t) \hspace{0.1em} e^t \hspace{0.2em}dt $$

For instance, with a polynomial function $f(t)$, we integrate many times in a row until the degree decreases to reach $0$.

We choose to integrate the exponential function so that the polynomial function is the one which is derivated and decrease in degree as integrations progresses.

Let us integrate this polynomial-exponential function:

$$ \gamma = \int^{x} (4t^3 - t +4 ) \hspace{0.1em} e^{2t} \hspace{0.2em}dt $$

$$ \Biggl \{ \begin{align*} f(t) = 4t^3 - t +4 \\ g'(t) = e^{2t} \hspace{0.2em}dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = (12t^2 -1) \hspace{0.2em}dt \\ g(t) = \frac{e^{2t}}{2} \end{align*} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - \int^{x} ( 12t^2 -1 ) \hspace{0.1em} \frac{e^{2t}}{2} \hspace{0.2em}dt $$

We take this oppotunity to take the constant out of it.

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6\int^{x} t^2 \hspace{0.1em} e^{2t} \hspace{0.2em}dt + \int^{x} \frac{e^{2t}}{2} \hspace{0.2em}dt $$

$$ \Biggl \{ \begin{align*} f(t) = t^2 \\ g'(t) = e^{2t} \hspace{0.2em}dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = 2t \hspace{0.2em}dt \\ g(t) = \frac{e^{2t}}{2} \end{align*} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6 \Biggl( \Biggl[ t^2 \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - \int^{x} t \hspace{0.1em} e^{2t} \hspace{0.2em}dt \Biggr) + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t} \Biggr]^{x} + 6 \int^{x} t \hspace{0.1em} e^{2t} \hspace{0.2em}dt + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \Biggl \{ \begin{align*} f(t) = t \\ g'(t) = e^{2t} \hspace{0.2em}dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = dt \\ g(t) = \frac{e^{2t}}{2} \end{align*} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t}\Biggr]^{x} + 6\Biggl[t \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6 \int^{x} \frac{e^{2t}}{2} \hspace{0.2em}dt + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t}\Biggr]^{x} + 3\Biggl[t \hspace{0.1em} e^{2t} \Biggr]^{x} - 3\Biggl[ \frac{e^{2t}}{2} \Biggr]^{x} + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \gamma = \frac{1}{2} \left (4x^3 - x +4 \right) \hspace{0.1em} e^{2x} - 3x^2 e^{2x} + 3x e^{2x} - \frac{5}{4}e^{2x} $$

We finally factorize it all.

$$ \gamma = e^{2x} \left (2x^3 - \frac{1}{2}x +2 - 3x^2 + 3x - \frac{5}{4} \right) \hspace{0.1em} $$

$$ \gamma = e^{2x} \left (2x^3 -3x^2 + \frac{5}{2} x + \frac{3}{4} \right) \hspace{0.1em} $$

Integration by substitution examples

Example 1

$$ A = \int_{0}^{\frac{\pi}{2}} \frac{sin(t)}{1 + 3 cos^2(t)} \ dt$$

In the case of a trigonometric polynomial function, we can apply Bioche's rules.

In our case, the integrand $ \omega(t) $ is invariant by the variable change $ t = -t $:

$$ \omega(-t) = \frac{sin(-t)}{1 + 3 cos^2(-t)} \ d(-t) $$

$$ \omega(-t) = \frac{-sin(t)}{1 + 3 cos^2(t)} \ -d(t) $$

$$ \omega(-t) = \frac{sin(t)}{1 + 3 cos^2(t)} \ -(t) $$

$$ \omega(-t) = \omega(t) $$

We thus set a new variable down:

$$ u = cos(t)$$

We do have now:

$$ \Biggl \{ \begin{align*} u = cos(t) \\ du = -sin(t) \ dt \end{align*} $$

We replace it in our main expression, without forgetting the bounds:

$$ A = \int_{cos(0)}^{cos(\frac{\pi}{2})} \frac{-du}{1 + 3 u^2} $$

$$ A = -\int_{1}^{0} \frac{1}{1 + 3 u^2} \ du $$

Thanks to this property of the integrals, we know that:

$$ \forall (a,b) \in I^2, \ \int_{b}^a f(t) \hspace{0.2em}dt = -\int_{a}^b f(t) \hspace{0.2em}dt $$

$$ A = \int_{0}^{1} \frac{1}{1 + 3 u^2} \ du$$

The antiderivative of the arctangent function is known as:

$$ \int^{x} \frac{1}{a^2 + t^2} \ dt = \Biggl[\frac{1}{a} arctan\left( \frac{t}{a}\right) \Biggr]^x $$

Alors, dans notre cas,

$$ A = \frac{1}{3} \int_{0}^{1} \frac{1}{\frac{1}{3} + u^2} \ du$$

Puis,

$$ A = \frac{1}{3} \int_{0}^{1} \frac{1}{ \left(\frac{1}{\sqrt{3}}\right)^2 + \ u^2} \ du = \frac{1}{3} \Bigl[\sqrt{3} \ arctan\left(\sqrt{3} u\right) \Bigr]_0^1 $$

$$ A = \frac{1}{3} \ \Bigl[ \sqrt{3} \ arctan(\sqrt{3}) - \sqrt{3} \ arctan(0) \Bigr] $$

$$ A = \frac{\sqrt{3}}{3} arctan(\sqrt{3}) $$

Example 2

$$ B = \int_{0}^{1} \ \frac{t^2}{1 + t^3} \ dt$$

The idea is to set $ u = t^3 $ down to get $ t^2 \ dt$ at the numerator (mutiplied by a constant) after substitution.

$$ \Biggl \{ \begin{align*} u = t^3 \\ du = 3t^2 \ dt \end{align*} $$

We replace it all (bounds do not change because $ 0$ and $ 1 $ do not vary taking their cube):

$$ B = \int_{0}^{1} \ \frac{1}{1 + u^3} \ \frac{du}{3} $$

$$ B = \frac{1}{3} \int_{0}^{1} \ \frac{1}{1 + u} \ du$$

$$ B = \frac{1}{3} \Bigl[ln |1 + u| \Bigr]_{0}^{\ 1} $$

$$ B = \frac{1}{3} ln(2) $$

Example 3

$$ C = \int_{1}^{3} \ \frac{e^{2t}}{1 - e^t} \ dt$$

The integrand in only defined for $ \mathbb{R} \ \backslash \{0 \}$.

$$ \Biggl \{ \begin{align*} u = e^t \\ du = e^t \ dt \end{align*} $$

$$ C = \int_{e}^{e^3} \ \frac{u}{1 - u} \ du$$

Let us use a little trick to transform the expression:

$$ C = \int_{e}^{e^3} \ \frac{u + 1 - 1}{1 - u} \ du$$

$$ C = \int_{e}^{e^3} \ \frac{u - 1}{1 - u} + \frac{1}{1 - u} \ du$$

$$ C = \int_{e}^{e^3} \ \frac{-(1-u)}{1 - u} + \frac{1}{1 - u} \ du$$

$$ C = -\int_{e}^{e^3} du + \int_{e}^{e^3} \frac{1}{1 - u} \ du$$

$$ C = - \Bigl[u\Bigr]_{e}^{e^3} + \Bigl[ - ln |1 - u| \Bigr]_{e}^{e^3} $$

$$ C = - e^3 + e - ln |1 - e^3| + ln |1 - e| $$

$$ C = - e^3 + e - ln (e^3-1) + ln(e-1) $$

Standard antiderivatives and general integration methods

Demonstrations

Recall on standard antiderivatives

General integration methods

Integration by parts

Integration by substitution

General integration methods recap table

Integration examples

Integration by parts examples

Integration by substitution examples