What is the integral of 1/(x^3)sin(1/(x^2)) ?

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

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integral of 1/(x^3)sin(1/(x^2))

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int - sin (v) d v

= \frac{1}{2} \cdot \int - sin (v) d v

Take the constant out:

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

= \frac{1}{2} (- \int sin (v) d v)

Use the common integral:

\int sin (v) d v = - cos (v)

= \frac{1}{2} (- (- cos (v)))

Substitute back

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

Simplify

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{3 n + 2}

\frac{d y}{d x} x^{2} y = y - 7

\int_{- 1}^{3} x^{2} - 2 x - 3 d x

\int \frac{6}{x ln ( 3 x )} d x

n = 1 \sum \infty \frac{5}{n n}

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