What is the integral of 1/(θ^2)sin(1/θ)cos(1/θ) ?

The integral of 1/(θ^2)sin(1/θ)cos(1/θ) is 1/4 cos(2/θ)+C

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integral of 1/(θ^2)sin(1/θ)cos(1/θ)

\int \frac{1}{θ ^{2}} sin (\frac{1}{θ}) cos (\frac{1}{θ}) d θ

\frac{1}{4} cos (\frac{2}{θ}) + C

Solution steps

\int \frac{1}{θ ^{2}} sin (\frac{1}{θ}) cos (\frac{1}{θ}) d θ

Rewrite using trig identities

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

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{2}{θ} )}{θ ^{2}} d θ

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = \frac{2}{θ}

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

Simplify

\frac{1}{2} (- \frac{1}{2} (- cos (\frac{2}{θ}))) : \frac{1}{4} cos (\frac{2}{θ})

= \frac{1}{4} cos (\frac{2}{θ})

Add a constant to the solution

= \frac{1}{4} cos (\frac{2}{θ}) + C

\frac{d}{d t} (e^{5 t})

y = e^{9 x}

\frac{d}{d t} (e^{t} cos (2 t))

x^{2} ln (x^{2})

What is the integral of 1/(θ^2)sin(1/θ)cos(1/θ) ?
The integral of 1/(θ^2)sin(1/θ)cos(1/θ) is 1/4 cos(2/θ)+C