To begin, we must first remember the formula for integration by parts:

When selecting which term to define as u and which to define as dv, keep in mind that the u we define should be easy to differentiate to get du, and the dv we define should be easy to integrate to get v. If we look at the integral we're evaluating, it appears either term could be easily differentiated or integrated, so let's define x/2 as u and sin(4x)dx as dv. Our next step is to find du and v, so let's differentiate u and integrate dv:

Now that we have everything we need for our integration by parts formula, we can simply plug in the values and simplify the expression:

Our integral has both a sine term and a cosine term, so our first step is going to be simplifying the expression to ensure it is in terms of only one trigonometric function. To do this, we're going to need to factor out a cos(x) from our (cos(x))^3 term, so that we can use a trig identity to replace the remaining (cos(x))^2 term:

Recall the following trig identity, which we can rearrange for (cos(x))^2 and then plug into our integral in the place of this term:

Now we can see that the only factor left not in terms of sine is our last one, (cos(x))dx. In order to express this factor in terms of sine, we must look ahead to the u substitution we're going to perform to evaluate the integral. If we define u as sin(x), our next step is to determine du, which we can see by taking a simple derivative of u is cos(x)dx. This means we can represent each factor of the integral in terms of u as follows:

After representing each factor in terms of u through u substitution, we can rewrite our integral in a more simple form, which we can then multiply out and evaluate:

Now our final step is to simply substitute sin(x) back in for u:

When we use the method of partial fractions, our first step is to factor the polynomial in the denominator to see what the terms are going to be in the denominators of each of our partial fractions. So first we factor the denominator of the equation we're integrating, and then we write the equation in terms of our partial fractions:

If we were to add our two partial fractions together, they would need a common denominator. As with any other fractions, we find our common denominator by multiplying the numerator and denominator of one fraction by the denominator of the other. Then we add these fractions with like denominators together. In this case, we have:

This whole term is still equal to the original equation we're integrating, and now they both have the same denominator, as shown below, so we know their numerators must be equal as well:

Our next step is to solve for A and B. We do this by evaluating our expression above for some x value that will cancel our A term out, solving for B, and then evaluating it again for some x value that will cancel our B term out, solving for A. If we set the factors that A and B are multiplied by equal to 0, we can see that our A term will be 0 when x= -2, and our B term will be 0 when x=3. So now we evaluate our expression at these two values of x and solve for the term that is not cancelled out:

Now that we know our values for A and B, we plug them back into our original partial fractions and find that our integral 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 

. 

For two functions and , the composite functions and  are not always equal to each other (though they can be). Therefore the indefinite integrals of  and  are not always equal to each other either, making the answer choice with the composite functions the correct answer, since it describes a mathematical phenomenon that is not a property of integrals.

If , then .

By evaluating this indefinite integral we find that 

This integral requires an application of the Integration By Parts Formula. However, a slight substitution needs to be made prior to completing the Integration By Parts.

For Integration By Parts:

Where "u" is a differentiable function, and "dv" is a function that can be integrated.

If we look at the integral: 

We notice that  into two functions, one of which would resemble the exponent in Euler's number (e). 

We can now make a substitution:

The integral now becomes transformed into:

This integral can now be solved by using Integration by Parts.

Let:

Now plugging in the original values for "g" results in one of the answer choices:

When dealing with the integral of trigonometric functions, sometimes it is helpful to rewrite these integrals in terms of "sin" and "cos". Let's try this:

Now, using "U-Substitution", this somewhat "Tricky" integral can be transformed into something very familiar.

Let:

The transformed integral will now look like:

Replacing the "u" with the substitution we made above results in:

This answer is a valid choice, however the answer choices do not seem to show this value. 

By using the following logarithmic rule, we can find a similar representation to this result:

So we can rewrite the above result as:

Using the fact that:

The final result is:

Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":

For:

So for our given integral, we can rewrite like so:

This can be rewritten as two separate integrals and solved using a simple substitution.

Solving each integral individually, we have:

Substituting this into the integral results in:

The other integral is solved the same way:

Substituting this into the integral results in:

Now combining these two statements together results in one of the answer choices:

This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.

But first a substitution needs to be made:

Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:

In General:

1. If "m" is odd, then we make the substitution , and we use the identity .

2. If "n" is odd, then we make the substitution , and we use the identity .

For our given problem statement we will use the first rule, and alter the integral like so:

Now we need to substitute back into v:

Now we need to substitute back into u, and rearrange to make it look like one of the answer choices: