Pope Cox
08/03/2023 · Primary School
\( \int _{}^{}x^{-1}\sin (1+\frac{1}{x}) d x \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the integral
\[
I = \int x^{-1} \sin\left(1 + \frac{1}{x}\right) \, dx,
\]
we can use a substitution to simplify the expression. Let's set
\[
u = 1 + \frac{1}{x}.
\]
Then, we differentiate \(u\) with respect to \(x\):
\[
\frac{du}{dx} = -\frac{1}{x^2} \implies du = -\frac{1}{x^2} \, dx \implies dx = -x^2 \, du.
\]
Next, we need to express \(x\) in terms of \(u\). From our substitution, we have:
\[
u - 1 = \frac{1}{x} \implies x = \frac{1}{u - 1}.
\]
Now, substituting \(x\) into \(dx\):
\[
dx = -\left(\frac{1}{u - 1}\right)^2 du = -\frac{1}{(u - 1)^2} du.
\]
Now we can substitute \(x^{-1}\) and \(dx\) into the integral:
\[
x^{-1} = u - 1,
\]
so we have:
\[
I = \int (u - 1) \sin(u) \left(-\frac{1}{(u - 1)^2}\right) du.
\]
This simplifies to:
\[
I = -\int \frac{(u - 1) \sin(u)}{(u - 1)^2} \, du = -\int \frac{\sin(u)}{u - 1} \, du.
\]
Now we can split the integral:
\[
I = -\int \frac{\sin(u)}{u - 1} \, du.
\]
This integral does not have a simple elementary form, but it can be expressed in terms of special functions or evaluated numerically for specific limits.
Thus, the integral
\[
\int x^{-1} \sin\left(1 + \frac{1}{x}\right) \, dx
\]
can be expressed as
\[
-\int \frac{\sin(u)}{u - 1} \, du + C,
\]
where \(C\) is the constant of integration.
In conclusion, the integral does not yield a simple closed form in terms of elementary functions, but it can be expressed in terms of the integral of \(\frac{\sin(u)}{u - 1}\).
Quick Answer
The integral \( \int x^{-1} \sin\left(1 + \frac{1}{x}\right) \, dx \) can be expressed as \( -\int \frac{\sin(u)}{u - 1} \, du + C \), where \(C\) is the constant of integration.
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