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.

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

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 **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 **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**.

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.

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|**.

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**.

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

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

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

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