##### Methods of integrations

Basic rules of integration or ** table of integration** help us solve simple problem when the integral is given on standard form. But when the problem is more complicated we need more sophisticated methods like:

**Integration by Partial Fraction Decomposition**

**Integration of using some Trigonometric Identities**

### Integration by Substitution

The substitution method also called the u-substitution is used when an integral contains some function and its derivative or when the integrated is a ** composite function**. When finding the derivative of a composite function we used

**and since the integral is the reverse of derivative then the u-substitution is the reverse of the chain rule.**

__the chain rule__We set u equal to the function and rewrite the integral in terms of the new variable u. In the end when we write the answer we should substitute back the u. We do this to make the integral easier to solve.

Let’s explain this step by step

We need to integrate $ \displaystyle \int{{f(g(x)){g}'(x)dx}}$

If we substitute $ \displaystyle u=g(x)$ then $ \displaystyle du={g}'(x)dx$ and $ \displaystyle dx=\frac{{du}}{{{g}'(x)}}$

Writing the integral in terms of the new variable u:

$ \displaystyle \int{{f(u)}}\cancel{{{g}'(x)}}\frac{{du}}{{\cancel{{{g}'(x)}}}}$

Then we get to solve the integral of some function we know:

$ \displaystyle \int{{f(u)du=F(u)+C}}$

In the end we substitute back the variable x at the answer:

$ \displaystyle F(u)+C=F(g(x)+C$

### Integration by Parts

Integration by parts method is used when we want to integrate the product of two functions. When finding the derivative of the product of two functions we use the ** the product rule**, and since the integral is the reverse of derivative then integration by parts is the reverse of the product rule.

Let’s explain this step by step

If we have two differentiable functions $ \displaystyle u(x)$ and $ \displaystyle v(x)$ and we want to find the derivative of their product we use __ the product rule__:

$\displaystyle (u\cdot v{)}’=u\cdot {v}’+v\cdot {u}’$ or written like $ \displaystyle d(uv)=udv+vdu$

By rearranging this rule we can write:

$ \displaystyle udv=d(uv)-vdu$

By integrating both sides we get:

$ \displaystyle \int{{udv=\int{{d(uv)-\int{{vdu}}}}}}$

Simplifying,we get the integration by part formula:

$ \displaystyle \int{{udv=uv-\int{{vdu}}}}$

### Integration by Partial Fraction

If the function that need to be integrated is in the form of an algebraic fraction which is not easy to evaluate then you need to write the fraction into partial fraction to make it simpler for integration. You have to remember that this method is done only if the degree of the numerator is less than the degree of the denominator.

But what if it’s not?!

We know that a rational function is a ratio of two polynomials $ \displaystyle \frac{{P(x)}}{{Q(x)}}$, where $ \displaystyle Q(x)\ne 0$

We have two types of partial fractions:

1. Proper Partial Fraction

When the degree of the denominator is less than the degree of the numerator, than the partial fraction is a proper partial fraction.

2. Improper Partial Fraction

When the degree of the numerator is greater than the degree of the denominator, than the partial fraction is an improper partial fraction. We can reduce the improper fraction into a proper one by using __the long division process__.

### Partial Fraction Decomposition

The process of taking a rational expression and decomposing it into simpler rational expression that we can add or subtract to get the original rational expression is called partial fraction decomposition.

### How to decompose an algebraic fraction into partial fraction?

We will have a rational expression on the form $ \displaystyle f(x)=\frac{{P(x)}}{{Q(x)}}$ where both $ \displaystyle P(x)$ and$ \displaystyle Q(x)\ne 0$ are polynomials and the degree of $ \displaystyle P(x)$ is smaller than the degree of $ \displaystyle Q(x)$.

After determining that it can turn into a partial fraction then we factor the denominator as completely as possible.

For each factor in the denominator we can use the table below:

**Integration of using some Trigonometric Identities**

__to simplify the functions so we can integrate them easily.__

**trigonometric identities**Some trigonometric identities you may encounter when solving integrals:

$ \displaystyle {{\sin }^{2}}x=\frac{{1-\cos 2x}}{2}$

$ \displaystyle {{\cos }^{2}}x=\frac{{1+\cos 2x}}{2}$

$ \displaystyle {{\sin }^{3}}x=\frac{{3\sin x-\sin 3x}}{4}$

$ \displaystyle {{\cos }^{3}}x=\frac{{3\cos x+\cos 3x}}{4}$

For more trigonometric identities check __trigonometric identities__

### Integration of Inverse Trigonometric Forms

We know the derivatives of the inverse trigonometric functions

$\displaystyle (\text{arcsinx}{)}’=\frac{1}{{\sqrt{{1-{{x}^{2}}}}}}$

$\displaystyle (\text{arccosx}{)}’=\frac{{-1}}{{\sqrt{{1-{{x}^{2}}}}}}$

$\displaystyle (\text{arctgx}{)}’=\frac{1}{{1+{{x}^{2}}}}$

Using those derivatives above, we can obtain the integrals as below, where u is a function of x that u=f(x).

$ \displaystyle \int{{\frac{{du}}{{\sqrt{{{{a}^{2}}-{{u}^{2}}}}}}}}=\arcsin \frac{u}{a}+C$

$ \displaystyle \int{{\frac{{du}}{{{{a}^{2}}+{{u}^{2}}}}}}=\frac{1}{a}\arctan \frac{u}{a}+C$

**Integration by Trigonometric Substitution**

Depending on the function we need to integrate, we can use this trigonometric expression as substitution to simplify the integral:

1. When $ \displaystyle \sqrt{{{{a}^{2}}-{{x}^{2}}}}$ then substitute $ \displaystyle x=a\sin \theta $

2. When $ \displaystyle \sqrt{{{{a}^{2}}+{{x}^{2}}}}$ then substitute $ \displaystyle x=a\tan \theta $

3. When $ \displaystyle \sqrt{{{{x}^{2}}-{{a}^{2}}}}$ then substitute $ \displaystyle x=a\sec \theta $