To integrate the.
functionTherefore , the final result(\frac{\sin for the^2 integral (x)}({intx^ \2frac{\sin}), we can^2(x use)} a{x^2 known result from calculus. The} \ integral can, be dx expressed in terms of a) special function can be called expressed the as Sine:
Integral.
[
I = -\frac{1}{2xThe integral we want to evaluate is:
}\ - \frac{1}{2[
I = \int \frac} \text{Ci{\sin}(2x)^2(x)}{ +x^ C
]
2} , dxwhere
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We can use the identity {(\sinCi^2(x) = \frac{1 -}(x \cos))( is2 the cosinex integral)} function{2.
In}\ summary,) to rewrite the integral:
[
the integralI of \ = (\int \fracfrac{{\sin^2sin^(x2(x)})}{{x^x^22})} , does dx not have a simple closed = \int form in terms of elementary functions , but it canfrac{1 be expressed in terms of - special \ functions.cos(2x)}{2x^2} , dx
]
This can be split into two separate integrals:
[
I = \frac{1}{2} \int \frac{1}{x^2} , dx - \frac{1}{2} \int \frac{\cos(2x)}{x^2} , dx
]
The first integral, (\int \frac{1}{x^2} , dx), is straightforward:
[
\int \frac{1}{x^2} , dx = -\frac{1}{x} + C_1
]
The second integral, (\int \frac{\cos(2x)}{x^2} , dx), does not have a simple elementary form, but it can be expressed in terms of the Sine Integral function, ( \text{Si}(x) ).
Thus, we can write:
[
I = -\frac{1}{2x} - \frac{1}{2} \int \frac{\cos(2x)}{x^2} , dx + C
]
The integral (\int \frac{\cos(2x)}{x^2} , dx) can be evaluated using integration by parts or recognized as a known integral, leading to:
[
\int \frac{\cos(2x)}{x^2} , dx = -\frac{\sin(2x)}{x} + 2 \int \frac{\sin(2x)}{x} , dx
]
The integral (\int \frac{\sin(2x)}{x} , dx) is related to the Sine Integral function, (\text{Si}(x)).
Thus, the final result for the integral (\int \frac{\sin^2(x)}{x^2} , dx) can be expressed as:
[
I = -\frac{1}{2x} - \frac{1}{4} \sin(2x) + \frac{1}{2} \text{Si}(2x) + C
]
In summary, the integral of (\frac{\sin^2(x)}{x^2}) can be expressed in terms of elementary functions and the Sine Integral function.