What is the integral of x^7*sin(x^4) ?

The integral of x^7*sin(x^4) is 1/4 (-x^4cos(x^4)+sin(x^4))+C

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integral of x^7*sin(x^4)

\int x^{7} \cdot sin (x^{4}) d x

\frac{1}{4} (- x^{4} cos (x^{4}) + sin (x^{4})) + C

Solution steps

\int x^{7} sin (x^{4}) d x

Apply u-substitution

: \int \frac{u sin ( u )}{4} d u

= \int \frac{u sin ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u sin (u) d u

Apply Integration By Parts

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

= \frac{1}{4} (- u cos (u) - \int - cos (u) d u)

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

= \frac{1}{4} (- u cos (u) - (- sin (u)))

Substitute back

u = x^{4}

= \frac{1}{4} (- x^{4} cos (x^{4}) - (- sin (x^{4})))

Simplify

= \frac{1}{4} (- x^{4} cos (x^{4}) + sin (x^{4}))

Add a constant to the solution

= \frac{1}{4} (- x^{4} cos (x^{4}) + sin (x^{4})) + C

f (x) = cos (x^{2})

n = 1 \sum \infty n (\frac{4}{3})^{n}

f (x) = 2 x - 3 x^{2}, at x = - 1

\frac{d y}{d t} = y + 2 t

- 19.6 t + 2

What is the integral of x^7*sin(x^4) ?
The integral of x^7*sin(x^4) is 1/4 (-x^4cos(x^4)+sin(x^4))+C