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

The integral of 1/(x^2)sin^2(1/x) is 1/2 (-1/x+1/2 sin(2/x))+C

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integral of 1/(x^2)sin^2(1/x)

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

\frac{1}{2} (- \frac{1}{x} + \frac{1}{2} sin (\frac{2}{x})) + C

Solution steps

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

Simplify

\frac{1}{x ^{2}} sin^{2} (\frac{1}{x}) : \frac{sin ^{2} ( \frac{1}{x} )}{x ^{2}}

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

Rewrite using trig identities

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

Take the constant out:

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

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

Expand

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

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

Apply the Sum Rule:

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

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

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

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

= \frac{1}{2} (- \frac{1}{x} - (- \frac{1}{2} sin (\frac{2}{x})))

Simplify

= \frac{1}{2} (- \frac{1}{x} + \frac{1}{2} sin (\frac{2}{x}))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{x} + \frac{1}{2} sin (\frac{2}{x})) + C

\int_{- 2}^{4} x - 9 d x

f (x) = 6 (x - 2)

\int_{- \infty}^{3} \frac{1}{x ^{2} + 1} d x

f (x) = x^{3} - 12 x, at x = 1

\frac{d}{d x} (\frac{x - 13}{x - 13})

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