# Getting two different integrals for same function(?)

• 1/2"
In summary, the conversation discusses solving the integral ∫sin2x/((sinx)4+(cosx)4) dx and the two different answers obtained using different methods. It is determined that the difference is a constant and with the given limits, there may be some doubt about the validity of the solutions. A suggestion is made to fix this by taking the integral 0 to π/4 and doubling it.
1/2"

## Homework Statement

The actual problem is ∫sin2x/((sinx)4+(cosx)4) dx

## The Attempt at a Solution

First wrote the expression as
∫$\frac{2sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }$ dx
then I changed the 2dx to d(2x)
∫$\frac{sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }$ d2x

∫$\frac{sin2x}{1+ (cos2x)^2 }$ d2x

Multiplying the whole expressioin with (sec2x)^2

∫$\frac{tan2x sec2x}{1+ (sec2x)^2 }$ d2x

tan2xsec2x d(2x) = d((sec2x)^2)

taking sec2x =t

∫$\frac{1}{1+ t^2 }$ dt
At this stage I am getting 2 different result when i proceed in 2 different way

First one:
integrating i get
arctan t = arctan (sec 2x)

Second one :
- ∫-$\frac{1}{1+ t^2 }$ dt

- arccot t

-arctan (cos2x)

Am i going wrong somewhere? please could you point it out ?

When an indefinite integral appears to give two different answers, it generally turns out that the difference is a constant. I believe that is the case here.

1 person
Actually it has got limits - 0 and $\frac{∏}{2}$

And putting the limits the answer in the first case come as $\frac{∏}{2}$ and in the second one -$\frac{∏}{2}$

haruspex said:
When an indefinite integral appears to give two different answers, it generally turns out that the difference is a constant. I believe that is the case here.

Ya the difference is a constant arctan ∞ = $\frac{∏}{2}$

If I happen to give either of the answer will it be marked correct ? or do I have to work out both of them ?

Last edited:
1/2" said:
Actually it has got limits - 0 and $\frac{∏}{2}$

And putting the limits the answer in the first case come as $\frac{∏}{2}$ and in the second one -$\frac{∏}{2}$
With those limits, there will be some doubt about the validity since your integral goes through an infinity at π/4. You could fix that by observing there's symmetry about π/4, so you can take the integral 0 to π/4 and double it.
However, I don't see why you are getting different answers. When I substitute those limits in your integrals I get $\frac{∏}{2}$ for both.

## 1. How is it possible to get two different integrals for the same function?

It is possible to get two different integrals for the same function because integration involves finding the antiderivative of a function, which may not be unique. This means that there can be multiple functions that have the same derivative, leading to different integrals.

## 2. Why do different methods of integration give different results for the same function?

Different methods of integration, such as substitution, integration by parts, or partial fractions, may yield different results because they use different techniques and strategies to find the antiderivative of a function. Additionally, the choice of method may depend on the complexity of the function being integrated.

## 3. How can I determine which integral is correct for a given function?

To determine which integral is correct for a given function, you can check your work by differentiating both integrals and seeing if they result in the original function. You can also compare your results to known formulas or use graphing software to visualize the functions.

## 4. Can two different integrals for the same function have the same value?

Yes, it is possible for two different integrals for the same function to have the same value. This can occur when the integrals represent different forms of the same function or when the functions being integrated are equal except for a constant term.

## 5. Are there any cases where two different integrals for the same function will always have the same value?

Yes, there are certain cases where two different integrals for the same function will always have the same value. This can occur when the function being integrated is an odd function, as the integral of an odd function over a symmetric interval will always be zero regardless of the method used.

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