# Standard Integrals

## Helpful Integrals to Simplify Tricky Problems

Solving tricky problems can be made easier by keeping a few key integrals in mind. These integrals not only help us streamline our work, but also provide a framework for solving similar problems.

### Trigonometric Integrals

Not all trigonometric integrals are simple, especially those involving basic trigonometric functions. Let's explore the integrals of some common trigonometric functions.

#### The Integral of Cosecant

Finding the integral of cosec (ax) dx, where a is a constant, may seem daunting at first. However, by using a simple trick we can make it much more manageable.

Multiplying the integral by -sin (ax), is equivalent to multiplying by 1. This gives us: ∫ -sin(ax)cosec(ax)dx.

Next, we use the substitution: u = sin (ax), which means du = a cos (ax) dx. Substituting this in the integral gives us: -∫ u du.

The solution is: -u^2 / 2 + C

Remember to add the integration constant at the end. So, the final answer is: -sin^2 (ax) / 2 + C.

#### The Integral of Secant

The integral of sec (ax) dx can also be solved using the same trick.

First, we define u = cos (ax), with a as the constant. Then, we multiply the integral by cos (ax), giving us: ∫ cos (ax) sec (ax) dx.

Substituting u into the integral, we get: ∫ u du. After integrating, the final answer is: cos^2 (ax) / 2 + C.

#### The Integral of Cotangent

The integral of cot (ax) dx requires a different approach and does not need an additional trick.

Defining u = tan (ax), with a as the constant, we can rewrite the integral as: ∫ 1 / u du.

Using the laws of logs, the solution is: ln |u| + C. Substituting u = tan (ax), the final answer is: ln |tan (ax)| + C.

### Useful Reciprocal Polynomial Integrals

When integrating polynomials, the solutions are usually straightforward. However, there are cases where the answers can seem to come out of nowhere.

As an example, let's solve the integral of 1 / (x^4 + 1). Though it may seem intimidating, it is actually quite simple to solve. We can factorize the denominator as: x^4 + 1 = (x^2 + 1)^2 - 2x^2.

Now, we use the substitution: t = x^2 + 1, which means: dt = 2x dx. Substituting this in the integral, we get: ∫ 1 / (t^2 - 2) dt.

After using partial fractions and solving for t, the solution is: 1 / 4 tan^(-1) (x^2) + C.

Similarly, the integral of 1 / (x^2 + 1) can be solved using two different methods, both yielding different but equivalent results.

#### Method 1:

First, we define t = x^2 + 1, giving us: ∫ 1 / t dt. Using the trigonometric identity: cot^(-1) (x) = -i ln |x + sqrt(1 + x^2) + C|.

#### Method 2:

Another method is to define t = x - i, giving us: ∫ 1 / t dt = ln |t| + C = ln |x^2 + 1| + C.

Though these methods may seem different, they can be verified as equivalent by using the identity: tan^(-1) (x) - cot^(-1) (x) = pi / 2.

### Standard Integrals - Key Takeaways

To make solving common integrals easier, it is helpful to remember the formulas for those that behave differently and come up frequently. Some standard integrals to keep in mind are:

• ∫ (cosec (x))^n dx = - (cosec (x))^(n-1) cot (x) / (n - 1) + C
• ∫ (sec (x))^n dx = (sec (x))^(n-1) tan (x) / (n - 1) + C
• ∫ (cot (x))^n dx = - (cot (x))^(n-1) csc (x) / (n - 1) + C
• ∫ (x^n / (x^2 + a^2)) dx = 1 / [(n - 1)(x^2 + a^2)^(n/2)] + C
• ∫ (x^n / (x^2 - a^2)) dx = 1 / [2(n + 1)(x^2 - a^2)^(n+1/2)] + C

#### What are Standard Integrals?

Standard integrals are those that differ from the norm and/or arise frequently, making it useful to remember their formulas.

#### How to Verify a Standard Integral?

As with any integral, we can differentiate the solution to ensure it leads back to the original function.