# Integration: Evaluate the Integral?

• KAISER91
In summary, the integral of x sec^2 x dx can be solved by using integration by parts and substitution. The correct answer is x tan x + ln(cos x) + c. This can also be expressed as x sin x / cos x + ln(sin x / cos x) + c or x sin x / cos x + ln(sec x) + c.
KAISER91

Evaluate;

∫ x sec^2 x dx

## The Attempt at a Solution

So far this is what I have;

u = x
du/dx = 1
du = 1 dx

dv = sec^2 x
v = tan x

Therefore;

= x tan x - ∫ tan x dx
= x tan x - ln (sec x) + c

I stopped there thinking that was the answer?

x tan x + ln (cos x) + c

Thanks

Where did you get that
$$\int tan(x)dx= ln(sec(x))+ C$$?

$$\int tan(x)dx= \int \frac{sin(x)}{cos(x)}dx$$
Let u= cos(x) so du= sin(x)
$$\int tan(x)dx= \int \frac{du}{u}= ln(u)+ C= ln(cos(x))+ C$$

Yeah, the answer given is correct. You must have just integrated tanx incorrectly. Hint: this integral is done easily by substitution

KAISER91 said:
Therefore;

= x tan x - ∫ tan x dx
= x tan x - ln (sec x) + c

I stopped there thinking that was the answer?

x tan x + ln (cos x) + c

What is another way of expressing ln (sec x)?

HallsofIvy said:
Where did you get that
$$\int tan(x)dx= ln(sec(x))+ C$$?

$$\int tan(x)dx= \int \frac{sin(x)}{cos(x)}dx$$
Let u= cos(x) so du= sin(x)
$$\int tan(x)dx= \int \frac{du}{u}= ln(u)+ C= ln(cos(x))+ C$$
OH.Thanks. I appreciate the help.

Note, you forgot the negative sign on sin x for the derivative of cos x.Thanks.

George Jones said:
What is another way of expressing ln (sec x)?
sin x / cos xLOLOL. I totally forgot about it.

KAISER91 said:
sin x / cos x

I'm not sure what you mean.

sec x = ?

ln(sec x) = ?

## Q1: What is integration and why is it important?

Integration is a mathematical concept that involves finding the area under a curve. It is important because it allows us to calculate the total amount of something, such as distance, volume, or energy, by breaking it down into smaller, more manageable parts.

## Q2: How do you evaluate an integral?

To evaluate an integral, you use a variety of techniques, such as substitution, integration by parts, or trigonometric identities. These methods involve manipulating the integral to make it easier to solve, and then using basic algebra and calculus rules to find the solution.

## Q3: What is the difference between definite and indefinite integration?

Definite integration involves finding the specific numerical value of an integral, while indefinite integration involves finding the general form of a function that, when differentiated, will result in the original function being integrated.

## Q4: What are some real-life applications of integration?

Integration has many practical applications, such as calculating the area of a region, finding the volume of a solid object, and determining the work done by a variable force. It is also used in physics, engineering, economics, and other fields to solve complex problems.

## Q5: Can integration be used to solve differential equations?

Yes, integration is an essential part of solving differential equations. It allows us to find the general solution of a differential equation, which is a function that satisfies the equation for all values of the independent variable. Integration is also used to find particular solutions to differential equations that satisfy given initial or boundary conditions.

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