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Example Questions
Example Question #1 : Indefinite Integrals
Find the indefinite integral of the following function:
To integrate this function, use u substitution. Make,
then substitute them into the equation to get
.
The integral of
then plug u back into the equation
.
The +C is essential because the integral is indefinite.
Example Question #1 : Indefinite Integrals
Evaluate the given indefinite integral
.
To integrate this function, use u substitution. Make
then substitute them into the equation to get
.
The integral of
then plug u back into the equation
.
The +C is essential because the integral is indefinite.
Example Question #2 : Indefinite Integrals
Evaluate the given indefinite integral
.
To integrate this function, use u substitution. Make
then substitute them into the equation to get
.
The integral of
so we have
.
The +C is essential because the integral is indefinite.
Example Question #1 : Indefinite Integrals
Calculate the following indefinite integral.
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
.
Example Question #2381 : Calculus Ii
If
What is
?
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.
Example Question #1 : Indefinite Integrals
Solve the following for .
Assume the integration constant is zero.
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
Example Question #1 : Indefinite Integrals
Solve the following for .
Assume the integration constant
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.
Example Question #2382 : Calculus Ii
Evaluate the following integral.
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
.
Example Question #9 : Indefinite Integrals
Find the indefinite integral of .
None of the above
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
.
Example Question #1 : Indefinite Integrals
Find the indefinite integral of .
None of the above
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
.
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