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

The integral of 1/(t^2)sin(2/1+3) is -(sin(5))/t+C

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integral of 1/(t^2)sin(2/1+3)

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

- \frac{sin ( 5 )}{t} + C

Solution steps

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

Take the constant out:

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

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

Simplify

= sin (5) \cdot \int \frac{1}{t ^{2}} d t

Apply the Power Rule

: - \frac{1}{t}

= sin (5) (- \frac{1}{t})

Simplify

sin (5) (- \frac{1}{t}) : - \frac{sin ( 5 )}{t}

= - \frac{sin ( 5 )}{t}

Add a constant to the solution

= - \frac{sin ( 5 )}{t} + C

\frac{d}{d x} (sec (x) an x)

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

\frac{\partial}{\partial y} (x^{2} + sin (x y))

\int_{0}^{1} cos (x) d x

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

What is the integral of 1/(t^2)sin(2/1+3) ?
The integral of 1/(t^2)sin(2/1+3) is -(sin(5))/t+C