### All Calculus 3 Resources

## Example Questions

### Example Question #41 : Integration

Evaluate the following integral:

**Possible Answers:**

**Correct answer:**

The only way to evaluate this integral is through integration by parts. To do this, you must follow the equation:

You must assign values for u and dv from the original integrand and then find the values of du and v. Specifically:

, , ,

From here, plug in the values into the equation for integrating by parts:

To evaluate this integral, a u-substitution is needed:

,

Now, the integrand from the equation can be rewritten as:

When you replace u with x and add the other half of the equation, the final answer is:

### Example Question #42 : Integration

Evaluate the following integral:

**Possible Answers:**

**Correct answer:**

The only way to evaluate this integral is through integration by parts where the formula states:

Values for u and dv must be picked from the integrand and then the remaining values are found from those. Specifically:

, , ,

Now, these values must be plugged into the equation:

The integral that now must be evaluated can be done so using a u-substitution:

,

Next, every x must be replaced with u and integrated:

After replacing u with x and adding the remainder of the equation, the final answer is:

### Example Question #43 : Integration

Evaluate the following integral:

**Possible Answers:**

**Correct answer:**

The only way to evaluate this integral is by splitting it in to two separate integrals:

The first integral can be evaluated by using a u-substitution:

,

Then, x must be replaced with u and evaluated:

For the second integral, the integrand resembles the form:

In this case, and the integral of it is

After adding all of the evaluated integrals, the final answer is:

### Example Question #44 : Integration

Evaluate the following integral:

**Possible Answers:**

**Correct answer:**

The only way to evaluate this integral is by recognizing that the integral resembles the following:

To evaluate the integral, the x-term must be replaced with u:

, ,

Now that the denominator is in the proper form, the integral can be evaluated according to the first equation:

U must be replaced with x to give the final answer:

### Example Question #152 : Calculus Review

Integrate:

**Possible Answers:**

**Correct answer:**

### Example Question #45 : Integration

Evaluate:

**Possible Answers:**

**Correct answer:**

Rewrite the function in terms of ln:

Plug in the high level and lower level limits:

Subtract higher from lower:

Answer=

### Example Question #154 : Calculus Review

Solve:

**Possible Answers:**

**Correct answer:**

The feature of this integral that may initially be confusing is the bounds of integration, because they contain a variable. However, this is something that will be used frequently in later integration problems. Treat the bounds as normal, and integrate:

After the definite integration, we get

### Example Question #46 : Integration

Integrate:

**Possible Answers:**

**Correct answer:**

To integrate, we must use integration by parts, which states that

We must designate a u and a dv from the integral given, and then differentiate and integrate, respectively:

,

Now, use the above formula and integrate again:

which simplified becomes

### Example Question #156 : Calculus Review

**Possible Answers:**

**Correct answer:**

This question requires the use of a complex conjugate. as the other methods of integration do not seem to work. In the case, we take the complex conjuage of the denominator, which is sec(x)+ 1:

Simplyifying the above equation results in:

We now notice that the denominator is a trig identity: , we can replace it with :

Now we can separate the above equation into two different integrals:

Simplifying each integral, now you get:

For the first integral, , we can use u-substitution to integrate it:

Crossing out cos(x), we now get:

Now integrate as normal:

Substitute u = sin(x)

Simplifying:

For the second integral, , we would use the trig identity . Rearranging the equation gets us :

Now we can integrate separately:

Now combining the result from the first integral, , we now get

### Example Question #157 : Calculus Review

**Possible Answers:**

**Correct answer:**

At first, we notice that the power in the numerator is smaller than the one in the denominator, but we can't quite use u-substitution yet, as the derivative of the denominator is 2t - not 2t -1. To do this we must separate the integrals:

Now, we can use u-substitution for the first integral by setting

Which results in:

The 2t's cross out and now we get:

Now integrate as normal:

Substitution back for , we now get:

For the second integral, , we have to use a special formula:

, or in this case translates to:

Combining the two results get us:

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