What is the integral of θ^3cos(θ^2) ?

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

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integral of θ^3cos(θ^2)

\int θ^{3} cos (θ^{2}) d θ

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

Solution steps

\int θ^{3} cos (θ^{2}) d θ

Apply u-substitution

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

= \int \frac{u cos ( 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 u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

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

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

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

Substitute back

u = θ^{2}

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

Simplify

= \frac{1}{2} (θ^{2} sin (θ^{2}) + cos (θ^{2}))

Add a constant to the solution

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

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

\frac{d y}{d t} = t y + t

\frac{d y}{d x} = \frac{1}{2} x + 4

\frac{d y}{d x} + x y = \frac{x}{y ^{3}}

\int \frac{7}{x ^{2} - 6 x + 25} d x

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