# A Integration Problem[Indefinite Integral]

• Hysteria X
In summary: Wait wait wait. Listen to what you're saying here. You said t = x2 right? So everywhere x2 appears in your equation you replace it with t.Now dt = 2x dx which implies that (1/2)dt = xdx. So anywhere you see xdx ( which it's VERY apparent to see ), you replace it with (1/2)dt.
Hysteria X

## Homework Statement

Evaluate

$$\int (x sin ( x^2 ) )/cos^3 ( x^2 ) dx$$

## Homework Equations

none the i know that would be helpful here

## The Attempt at a Solution

I tried solving this by substitution method by substituting $$t = x^2$$ as t but i don't know how to proceed after that

Hysteria X said:

## Homework Statement

Evaluate

$$\int (x sin ( x^2 ) )/cos^3 ( x^2 ) dx$$

## Homework Equations

none the i know that would be helpful here

## The Attempt at a Solution

I tried solving this by substitution method by substituting $$t = x^2$$ as t but i don't know how to proceed after that

Show us what you got, because you should be able to push that through. Alternatively, try ##t=\cos(x^2)##.

LCKurtz said:
Show us what you got, because you should be able to push that through. Alternatively, try ##t=\cos(x^2)##.

Damn how could i not see that
okay i got the answer as ##I= 1/4(1/cos^2(x^2)##

Can anyone confirm this

You don't need someone else to confirm it. Differentiate your answer and see if it works.

Hysteria X said:
Damn how could i not see that
okay i got the answer as ##I= 1/4(1/cos^2(x^2)##

Can anyone confirm this
Don't forget the constant of integration.

If you had gone with your first idea, $t= x^2$ so that dt= 2x dx and (1/2)dt= x dx, you would have got $(1/2)\int sin(t)/cos^3(t) dt$. Now, it should be clear, with that "sin(t)dt" in the numerator, you can let y= cos(t), so that dy= sin(t) dt, and the integral becomes $(1/2)\int dy/y^3= (1/2)\int y^{-3}dy$.

Essentially, you are doing the same thing in two steps rather than one.

HallsofIvy said:
If you had gone with your first idea, $t= x^2$ so that dt= 2x dx and (1/2)dt= x dx, you would have got $(1/2)\int sin(t)/cos^3(t) dt$. Now, it should be clear, with that "sin(t)dt" in the numerator, you can let y= cos(t), so that dy= sin(t) dt, and the integral becomes $(1/2)\int dy/y^3= (1/2)\int y^{-3}dy$.

Essentially, you are doing the same thing in two steps rather than one.

Sir correct me if i am wrong but i think there is a mistake in the method you have done which i first stated.

if we took $t= x^2$ and dt =2x dx we cannot substitute that in the equation because ##x^2## is within the sine function and dx is outside?

Hysteria X said:
Sir correct me if i am wrong but i think there is a mistake in the method you have done which i first stated.

if we took $t= x^2$ and dt =2x dx we cannot substitute that in the equation because ##x^2## is within the sine function and dx is outside?

Wait wait wait. Listen to what you're saying here. You said t = x2 right? So everywhere x2 appears in your equation you replace it with t.

Now dt = 2x dx which implies that (1/2)dt = xdx. So anywhere you see xdx ( which it's VERY apparent to see ), you replace it with (1/2)dt.

Clear now?

You replace all occurrences of x separately.

So the sine and cosine contain an x², which you substitute by t.

Then outside the trig functions, you are left with an x dx which - as you said - you substitute by (1/2) dt.

## 1. What is an indefinite integral?

An indefinite integral is a mathematical concept that represents the antiderivative of a function. It is denoted by ∫f(x)dx, where f(x) is the original function and dx represents the variable of integration. Essentially, it is the process of finding a function whose derivative is equal to the original function.

## 2. How is an indefinite integral different from a definite integral?

An indefinite integral represents a family of functions, while a definite integral represents a single numerical value. In a definite integral, the limits of integration are specified, whereas in an indefinite integral, they are not.

## 3. What are some techniques for solving indefinite integrals?

Some common techniques for solving indefinite integrals include the power rule, substitution, integration by parts, and partial fractions. These techniques can be used depending on the form of the original function and its complexity.

## 4. How do I know if my answer to an indefinite integral is correct?

You can check your answer by taking the derivative of the antiderivative you found. If the resulting function is equal to the original function, then your answer is correct. Additionally, you can also use online tools or a graphing calculator to verify your answer.

## 5. In what real-life situations would I need to use indefinite integrals?

Indefinite integrals are used in various fields of science, including physics, engineering, and economics. They are particularly helpful in calculating displacement, velocity, and acceleration, as well as in finding the area under a curve. In everyday life, they can also be used to solve problems involving rates of change, such as finding the average rate of growth of a population over time.

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