What is the integral of x^{39}sin(x^{40}) ?

The integral of x^{39}sin(x^{40}) is -1/40 cos(x^{40})+C

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integral of x^{39}sin(x^{40})

\int x^{39} sin (x^{40}) d x

- \frac{1}{40} cos (x^{40}) + C

Solution steps

\int x^{39} sin (x^{40}) d x

Apply u-substitution

: \int \frac{sin ( u )}{40} d u

= \int \frac{sin ( u )}{40} d u

Take the constant out:

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

= \frac{1}{40} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{40} (- cos (u))

Substitute back

u = x^{40}

= \frac{1}{40} (- cos (x^{40}))

Simplify

= - \frac{1}{40} cos (x^{40})

Add a constant to the solution

= - \frac{1}{40} cos (x^{40}) + C

x \to 6 lim (\frac{1}{( x - 6 ) ^{2}})

t \to π lim (cos (\frac{t}{4}))

y = x a^{2} - x^{2}

x - 2 sin (x)

y = x^{3}, y = 1, x = 2, x = 2

What is the integral of x^{39}sin(x^{40}) ?
The integral of x^{39}sin(x^{40}) is -1/40 cos(x^{40})+C