What is the integral of sin(-5x^4-x-6)(-20x^3-1) ?

The integral of sin(-5x^4-x-6)(-20x^3-1) is -cos(6+x+5x^4)+C

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integral of sin(-5x^4-x-6)(-20x^3-1)

\int sin (- 5 x^{4} - x - 6) (- 20 x^{3} - 1) d x

- cos (6 + x + 5 x^{4}) + C

Solution steps

\int sin (- 5 x^{4} - x - 6) (- 20 x^{3} - 1) d x

Apply u-substitution

: \int sin (u) d u

= \int sin (u) d u

Use the common integral:

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

= - cos (u)

Substitute back

u = - 5 x^{4} - x - 6

= - cos (- 5 x^{4} - x - 6)

Use the negative angle identity:

cos (- x) = cos (x)

= - cos (6 + x + 5 x^{4})

Add a constant to the solution

= - cos (6 + x + 5 x^{4}) + C

\int \frac{x}{1 - 3 x ^{2}} d x

f (x) = x^{2}, g (x) = 9 + x

f (x) = 1 8^{x}

\int \frac{y}{y ^{2} + 2} d y

\int (10 x - 3) d x

What is the integral of sin(-5x^4-x-6)(-20x^3-1) ?
The integral of sin(-5x^4-x-6)(-20x^3-1) is -cos(6+x+5x^4)+C