To calculate the integral, we need to use integration by parts. The definition for integration by parts is

It is important here to select the correct u and dv terms from our orginal integral. We eventually want of the terms to "go away" when we take its derivate. We notice here that out of our two functions in our integral,  and , the derivate of x is 1, making is very simple to integrate eventually. Therefore,  will be our  term, and  will be our dv term. Note that the dv term is not just dx, but the function attached to it as well. If  was our  term, then  would be our dv term.

Now calculate the terms  and  needed to proceed with the integration by parts equation.

(you can to set integration constant c=0)

Now that we have the terms that we need, we can plug in these terms into the integration by parts formula above.

 - 

Note that although we still need to integrate one more time, this new integral only consists of one function which is simple to integrate, as opposed to the two functions we had before. Also note that the x term from the initial integral "went away", thus making the resulting integral easy to calculate.

Simplifying this term now becomes

.

Although the integral looks difficult, it can be majorly simplified. Remember this crucial trig identity.

Using this identity, the integral can now be simplified to

, which is very simple to integrate.

In this problem we can try to get all the terms with  on one side and all the terms with  on the other.

Now we can integrate both sides using the definition of  and the power rule.

 and 

Move all expressions with  to one side and all  to the other.

Now integrate both sides using the power rule and the definition of natural log.

The power rule for integrals states, 

 and the definition of natural log is,

.

Applying these rules we are able to solve the problem.

We need to use the following identity:

Our integral now becomes

.

Notice inside the cosine becomes 4 because we already had a 2 in the original expression.

This can be split into two integrals

.

Which becomes 

.

We can find the indefinite integral of   using the Power Rule for Integrals, which states that 

 for all  and with the arbitrary constant of integration .

Applying this rule to 

. 

We can find the indefinite integral of   using the Power Rule for Integrals, which states that 

 for all  and with the arbitrary constant of integration .

Applying this rule to 

. 

Since doesn't resemble any basic integral form, we can use Integration by Parts, which follows this pattern.

Where is representing .

In general, we choose "u" to be the part of the integral that will eventually turn to zero if we differentiate it repeatedly, and "dv" to be the rest of the integral. In this case,

, and .

Now we differentiate "u" to find "du", and integrate "dv" to find "v". Doing so gives,

, and .

Now we plug the pieces into the Integration by Parts pattern.

Simplifying a little gives the following.

.

Unfortunately, this remaining integral, , is still not in a form that we can integrate. However, if we apply Integration by Parts to this second integral, we will get something that we can integrate. Using the same reasoning as before, choose "u" and "dv".

Then find "du" and "v" by the same method as before.

Now, put all the pieces together for the second iteration of Integration by Parts.

Now we have an integral that resembles a basic integral form. Let's integrate it.

Let's simplify this result before substituting it back into the first Integration by parts equation.

Now substitute this into the first Integration by parts result.

Now we simplify.

Thus, the final answer is

Neither nor will reduce to zero no matter how many times we differentiate them. So we must use Integration by Parts twice and solve for the integral algebraicly to find the answer. The process starts by choosing "u" and "dv". In this case, both and can be both differentiated and integrated without issue, so it doesn't matter which is "u" and which is "dv". In this example, we will choose the following.

Then we find "du" by differentiating "u", and we find "v" by integrating "dv".

Now we assemble these pieces into the Integration by Parts pattern.

Let's simplify a little and pull the constant, 2, out of the new integral.

Now we have to use Integration by parts on the new integral, , with "u" and "dv" chosen similarly to the first Integration by Parts.

Now differentiate "u" and integrate "dv".

Now apply Integration by parts to .

Lets simplify the new integral.

Substituting this back into the first Integration by Parts equation, we get

Now simplify by mulitplying the -2 throught the substituted expression.

Now notice that the original integral we are evaluating is now present on both sides of the equation. Solve for it by adding to both sides, thus canceling the right sides's, then combine them on the left side.

Now isolate the integral by dividing both sides by 5. This cancels the coefficient of 5, and leaves the integral that we are evaluating isolated on the left side of the equation.

We added the constant of integration into our final answer because this is an indefinite integral. This gives the final answer.

In order to integrate, we must use integration by parts, which follows the formula

We must set our  to be a function easy to differentiate:

This makes the integration step very easy later on!

The derivative was found using the following rule:

The rest of our given integrand is :

The integral was found using the following rule:

(We substitute

into the integral and integrate according to the above rule. Finally we replace  with our  term from above.)

Now, write out the integration by parts:

Finally, integrate:

The integration was performed using the following rule:

Our final answer is henceforth