Rules And Methods Of Integration

We got introduced to the concept of  earlier. Here we will learn about the various rules that apply to integration and some of the fundamental methods to integrate various functions.

I. Rules Of Integration
The following rules apply to integrations and are commonly used for solving integration problems.

$\displaystyle \int k \cdot f(x)dx = k \int f(x)dx$ for any constant $k$

$\displaystyle \int [f(x) \pm g(x)]dx = \int f(x)dx \pm \int g(x)dx$

$\displaystyle \int f(x) g(x) dx = \int g(x) f(x) dx $

The following sections explain different methods of finding integrals.

II. Integration Using Anti-Derivatives
We saw earlier, that integration is an inverse operation of differentiation. Therefore, if the given problem can be mapped to a well know derivation problem then it can be solved directly using anti-derivatives.
For example, if we were to find the integration:
$\displaystyle \int x^m dx$

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We know that $\dfrac{d}{dx} x^n = nx^{n-1}$
Substituting $m = n - 1$ in the above equation, we get:

$\dfrac{d}{dx} x^{m + 1} = (m + 1)x^{m}$

$\Rightarrow \dfrac{1}{m + 1} \dfrac{d}{dx} x^{m + 1} = x^m$

Integrating both sides w.r.t. $x$, we get:
$\displaystyle \int \left[\dfrac{1}{(m + 1)} \dfrac{d}{dx} x^{m + 1}\right] = \int x^m dx$

$\Rightarrow \dfrac{1}{(m + 1)} \displaystyle \int \left[ \dfrac{d}{dx} x^{m + 1}\right] = \int x^m dx$

Since, integration is an inverse operation of differentiation,
$\displaystyle \int \left[ \dfrac{d}{dx} x^{m + 1}\right] = x^{m+1}$

Therefore, we get:
$\Rightarrow \displaystyle \dfrac{1}{(m + 1)}  \left(x^{m + 1}\right) = \int x^m dx$

$\Rightarrow \displaystyle \int x^m dx = \dfrac{x^{m + 1}}{(m + 1)}$

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III. Integration By Substitution
In this approach, if the function that needs to be integrated does not have a direct known anti-derivative form, in some cases, substituting independent variable function form can help solve the problem.
That is if we are finding $\displaystyle \int f(x)dx$ where $f(x)$ is not a well-known anti-derivative, then we can investigate substituting:
$x = g(z)$ and perform the following steps.
$\dfrac{dx}{dz} = g'(z)$
$\Rightarrow dx = g'(z)dz$
Substituting this in our original integration problem, we get:
$\displaystyle \int f(x)dx$
$\displaystyle \int f(g(z)) g'(z)dz$
If this form can be integrated as:
$\displaystyle \int f(g(z)) g'(z)dz = F(z)$ we can obtain the final result by substituting $z = g^{-1}(x)$ and get the final result as:
$F(g^{-1}(x))$

As an example take a look at the following integration:
$\displaystyle \int x \sqrt{x - 10}dx$ 
If we substitute $\sqrt{x - 10} = z$, then
$z^2 = x - 10 \Rightarrow x = z^2 + 10$

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Differentiating both sides w.r.t $z$,
$2z = \dfrac{dx}{dz}$

$\Rightarrow dx = 2z \cdot dz$

Making our substitutions in the original problem, we get:
$\displaystyle \int x \sqrt{x - 10}dx$

$= \displaystyle \int ((z^2 + 10)z)(2z)dz$
$= \displaystyle \int (2z^4 + 20z^2)dz$
$= \displaystyle \int 2z^4dz + \int 20z^2dz$
$= \displaystyle 2\int z^4dz + 20 \int z^2dz$
$= 2 \times \dfrac{z^5}{5} + 20 \dfrac{z^3}{3} + C$

Substituting $z = \sqrt{x^2 - 1}$ in the above function, we get the final result as:
$= 2 \times \dfrac{(\sqrt{x^2 - 1})^5}{5} + 20 \dfrac{(\sqrt{x^2 - 1})^3}{3} + C$

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IV. Integration By Parts
If we were to find $\displaystyle \int u(x)\cdot v(x)dx$, we can do so by substituting:
$\displaystyle g(x) = \int v(x)dx$ and
$f(x) = u(x)$, then solving for the expression:
$\displaystyle u(x)v(x) - \int f(x)g(x) dx$

Before we proceed with our examples, we will see the derivation of this technique. This is a direct consequence of the product rule for  .
We know from the product rule, that:
$\dfrac{d}{dx} f(x)g(x) = f'(x)g(x) + f(x)g'(x)$
Integrating both sides w.r.t. $x$, we get
$\displaystyle f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx$
$\Rightarrow \displaystyle \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$

Now, reverting back our substitutions and replacing $f(x) = u(x)$, $g'(x) = v(x)$, that is, $g(x) = \displaystyle \int v(x)dx$, we get:
$\displaystyle \int u(x)v(x)dx = u(x)\int v(x)dx - \int \left[u'(x) \cdot \int v(x)dx\right]dx$

Here the main trick is to figure out that of the two functions, $u(x)$ and $v(x)$, which one to integrate and which one to differentiate.
The rule of thumb is to choose the function, which is simplified by differentiating as $u(x)$ and the other one as $v(x)$.

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As an example, we will find the integral of $x^2\ln x$ w.r.t $x$,
We know that $\dfrac{d}{dx} \ln x = \dfrac{1}{x}$, therefore it can cancel out with the power of $x$ from the other function.
Therefore, we will take $u(x) = \ln x$ and $v(x) = x^2$

$\displaystyle \int x^2\ln x dx$

$= \displaystyle \int (\ln x) x^2 dx$

$= \displaystyle \ln x \int x^2 dx - \int  \left[\dfrac{d}{dx}(\ln x) \times \int x^2 dx\right] dx$

$= \displaystyle \ln x \int x^2 dx - \int  \dfrac{1}{x} \times \dfrac{x^3}{3} dx$

$= \displaystyle \ln x \times \dfrac{x^3}{3} - \int \dfrac{x^2}{3} dx$

$= \displaystyle \dfrac{x^3 \ln x}{3} - \dfrac{x^3}{9} + C$

(We will expect the reader to verify this by taking the derivative of the result and matching it with the original problem)

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We will see one more example before closing this chapter.
Find the value of $\displaystyle \int 3^x \cos x dx$

Let $I = \displaystyle \int 3^x \cos x dx$
We know the differential of $a^x$ from    and integrating $\cos x$ or $\sin x$ will give us only $\sin x$ and $\cos x$
Therefore, we take $u(x) = 3^x$ and $v(x) = \cos x$ and we get:
$I = \displaystyle 3^x \int \cos xdx - \int \left[ \dfrac{d}{dx} 3^x \right] \left[ \int \cos x dx \right] dx$

$\Rightarrow I = \displaystyle 3^x \sin x - \int \left[ 3^x \ln 3 \right] \left[ \sin x \right] dx$

$\Rightarrow I = \displaystyle 3^x \sin x - (\ln 3)\int 3^x \sin x dx$        $...(i)$

Now, we will evaluate $\int 3^x \sin x dx$ separately, again using integration by parts, using the same function substitution as above:

$\int 3^x \sin x dx$

$= \displaystyle 3^x \int \sin x dx - \int \left[ \dfrac{d}{dx} 3^x \right] \left[ \int \sin x dx \right] dx$

$= \displaystyle -3^x \cos x + \ln 3\int \left[ 3^x \cos x \right]dx$

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Using our original variable assignment $\displaystyle I = \int 3^x \cos x dx$, we get:

$\displaystyle \int 3^x \sin x dx = \displaystyle -3^x \cos x + I \cdot \ln 3$

Substituting the above value in $eqn\ (i)$, we get:

$I = \displaystyle 3^x \sin x - (\ln 3) \left( -3^x \cos x + I \cdot \ln 3 \right)$

$\Rightarrow I = \displaystyle 3^x \sin x + 3^x \ln 3 \cos x - I \cdot (\ln 3)^2$

$\Rightarrow I + I \cdot (\ln 3)^2 = 3^x (\sin x + \ln 3 \cos x)$

$\Rightarrow I = \dfrac{3^x (\sin x + \ln 3 \cos x)}{1 + (\ln 3)^2} + C$