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}} \cdot 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

\int \frac{t}{t ^{2} + 9} d t

\int \frac{y ^{3} + 1}{y} d y

\int (x + y)^{- 2} d x

\int \frac{x ^{2} + 1}{x + 1} d x

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