What is the integral of 3/(x^2)sin^2(3/x) ?

The integral of 3/(x^2)sin^2(3/x) is 3/2 (-1/x+1/6 sin(6/x))+C

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integral of 3/(x^2)sin^2(3/x)

\int \frac{3}{x ^{2}} sin^{2} (\frac{3}{x}) d x

\frac{3}{2} (- \frac{1}{x} + \frac{1}{6} sin (\frac{6}{x})) + C

Solution steps

\int \frac{3}{x ^{2}} sin^{2} (\frac{3}{x}) d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2}} sin^{2} (\frac{3}{x}) d x

Simplify

\frac{1}{x ^{2}} sin^{2} (\frac{3}{x}) : \frac{sin ^{2} ( \frac{3}{x} )}{x ^{2}}

= 3 \cdot \int \frac{sin ^{2} ( \frac{3}{x} )}{x ^{2}} d x

Rewrite using trig identities

= 3 \cdot \int \frac{1 - cos ( \frac{6}{x} )}{2 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1 - cos ( \frac{6}{x} )}{x ^{2}} d x

Expand

\frac{1 - cos ( \frac{6}{x} )}{x ^{2}} : \frac{1}{x ^{2}} - \frac{cos ( \frac{6}{x} )}{x ^{2}}

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{x ^{2}} - \frac{cos ( \frac{6}{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

= 3 \cdot \frac{1}{2} (\int \frac{1}{x ^{2}} d x - \int \frac{cos ( \frac{6}{x} )}{x ^{2}} d x)

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

\int \frac{cos ( \frac{6}{x} )}{x ^{2}} d x = - \frac{1}{6} sin (\frac{6}{x})

= 3 \cdot \frac{1}{2} (- \frac{1}{x} - (- \frac{1}{6} sin (\frac{6}{x})))

Simplify

3 \cdot \frac{1}{2} (- \frac{1}{x} - (- \frac{1}{6} sin (\frac{6}{x}))) : \frac{3}{2} (- \frac{1}{x} + \frac{1}{6} sin (\frac{6}{x}))

= \frac{3}{2} (- \frac{1}{x} + \frac{1}{6} sin (\frac{6}{x}))

Add a constant to the solution

= \frac{3}{2} (- \frac{1}{x} + \frac{1}{6} sin (\frac{6}{x})) + C

What is the integral of 3/(x^2)sin^2(3/x) ?
The integral of 3/(x^2)sin^2(3/x) is 3/2 (-1/x+1/6 sin(6/x))+C