What is the integral of sin^2(1/x) ?

The integral of sin^2(1/x) is x-xcos^2(1/x)-\Si(2*1/x)+C

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integral of sin^2(1/x)

\int sin^{2} (\frac{1}{x}) d x

x - x cos^{2} (\frac{1}{x}) - Si (2 \cdot \frac{1}{x}) + C

Solution steps

\int sin^{2} (\frac{1}{x}) d x

Apply u-substitution

= \int - \frac{sin ^{2} ( u )}{u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{sin ^{2} ( u )}{u ^{2}} d u

Rewrite using trig identities

= - \int \frac{1 - cos ^{2} ( u )}{u ^{2}} d u

Expand

\frac{1 - cos ^{2} ( u )}{u ^{2}} : \frac{1}{u ^{2}} - \frac{cos ^{2} ( u )}{u ^{2}}

= - \int \frac{1}{u ^{2}} - \frac{cos ^{2} ( u )}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int \frac{1}{u ^{2}} d u - \int \frac{cos ^{2} ( u )}{u ^{2}} d u)

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{cos ^{2} ( u )}{u ^{2}} d u = - \frac{cos ^{2} ( u )}{u} - Si (2 u)

= - (- \frac{1}{u} - (- \frac{cos ^{2} ( u )}{u} - Si (2 u)))

Substitute back

u = \frac{1}{x}

= - (- \frac{1}{\frac{1}{x}} - (- \frac{cos ^{2} ( \frac{1}{x} )}{\frac{1}{x}} - Si (2 \cdot \frac{1}{x})))

Simplify

- (- \frac{1}{\frac{1}{x}} - (- \frac{cos ^{2} ( \frac{1}{x} )}{\frac{1}{x}} - Si (2 \cdot \frac{1}{x}))) : x - x cos^{2} (\frac{1}{x}) - Si (2 \cdot \frac{1}{x})

= x - x cos^{2} (\frac{1}{x}) - Si (2 \cdot \frac{1}{x})

Add a constant to the solution

= x - x cos^{2} (\frac{1}{x}) - Si (2 \cdot \frac{1}{x}) + C

What is the integral of sin^2(1/x) ?
The integral of sin^2(1/x) is x-xcos^2(1/x)-\Si(2*1/x)+C