Basic indefinite integral

In summary: Thanks for the input!In summary, this conversation is about the integrals arcsinh and arcosh. The first step is to u-substitute and then integrate by parts. The u-sub is simple, so I'll skip that part. The second step is to use the wrong inspection method and get the wrong answer. Thanks for the help everyone.
  • #1
Hi guys, I don't even know where to begin with this question.

Find the following indefinite integral:

[tex] \int x.arsinh (x^2) dx [/tex]

Thanks very much for any help, its much appreciated.
 
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  • #2
Well, first you'll need a u-substitution and then integration by parts. The u-sub is simple, so I'll skip that part. For the integration by parts, try putting u=arcsin(x) and dv=dx.
 
  • #3
Thanks for that, I got the following final answer:

[tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?
 
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  • #5
Electrifying said:
Thanks for that, I got the following final answer:

[tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?
The substitution should be u = arcsinh(x). There is no arsinh function.
 
  • #6
Woops! I thought that read arcsin(x)! My bad.
 
  • #7
Stimpon said:

Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been taught to write inverse hyperbolic trig functions as arsinh, arcosh etc. rather than arc. Maybe its just an English convention, or did my sixth form teach me wrong?
 
  • #8
Electrifying said:
Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been taught to write inverse hyperbolic trig functions as arsinh, arcosh etc. rather than arc. Maybe its just an English convention, or did my sixth form teach me wrong?
Did your teacher also write arcsin and arccos as arsin and arcos?
 
  • #9
No they were written as arcsin and arccos. It was just the hyperbolic ones that were written without a c.
 
  • #10
arsinh and arcosh are the correct forms; they stand for "area hyperbolic sine" and "area hyperbolic cosine", respectively.
 
  • #11
Char. Limit said:
arsinh and arcosh are the correct forms; they stand for "area hyperbolic sine" and "area hyperbolic cosine", respectively.
Well, you learn something every day.
 

What is a basic indefinite integral?

A basic indefinite integral is the inverse operation of differentiation. It is a mathematical concept that involves finding the most general antiderivative of a given function.

How is a basic indefinite integral represented?

A basic indefinite integral is represented using the integral symbol (∫) followed by the function to be integrated and the variable of integration.

What is the process for solving a basic indefinite integral?

The process for solving a basic indefinite integral involves applying the power rule, sum rule, and constant multiple rule to the given function. The resulting antiderivative is then expressed as an indefinite integral.

What is the difference between a basic indefinite integral and a definite integral?

A basic indefinite integral does not have upper and lower limits, while a definite integral has specific values for the upper and lower limits. This means a basic indefinite integral represents a family of functions, while a definite integral represents a single value.

Why are basic indefinite integrals important in mathematics?

Basic indefinite integrals are important in mathematics because they help us find the area under the curve of a given function, which has numerous real-world applications in fields such as physics, engineering, and economics. They also provide a fundamental understanding of the relationship between a function and its derivative.

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