What is the integral of 1/(t^3)cos(pi/(t^2)) ?

The integral of 1/(t^3)cos(pi/(t^2)) is -1/(2pi)sin(pi/(t^2))+C

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integral of 1/(t^3)cos(pi/(t^2))

\int \frac{1}{t ^{3}} cos (\frac{π}{t ^{2}}) d t

- \frac{1}{2 π} sin (\frac{π}{t ^{2}}) + C

Solution steps

\int \frac{1}{t ^{3}} cos (\frac{π}{t ^{2}}) d t

Simplify

\frac{1}{t ^{3}} cos (\frac{π}{t ^{2}}) : \frac{cos ( \frac{π}{t ^{2}} )}{t ^{3}}

= \int \frac{cos ( \frac{π}{t ^{2}} )}{t ^{3}} d t

Apply u-substitution

: \int \frac{cos ( \frac{π}{u} )}{2 u ^{2}} d u

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

Apply u-substitution

: \int - \frac{cos ( v )}{π} d v

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} (- \frac{1}{π} sin (v))

Substitute back

= \frac{1}{2} (- \frac{1}{π} sin (\frac{π}{t ^{2}}))

Simplify

\frac{1}{2} (- \frac{1}{π} sin (\frac{π}{t ^{2}})) : - \frac{1}{2 π} sin (\frac{π}{t ^{2}})

= - \frac{1}{2 π} sin (\frac{π}{t ^{2}})

Add a constant to the solution

= - \frac{1}{2 π} sin (\frac{π}{t ^{2}}) + C

y^{^{''}} - 4 y = e^{- 2 x} - 2 x, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = \frac{3}{2} sin (x)

\frac{x}{3} (2 x + 1)^{3}

(- 2, 5), (6, 2)

\int x + 1 x d x

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