What is the integral of (-21x^2-4x)cos(-7x^3-2x^2-2) ?

The integral of (-21x^2-4x)cos(-7x^3-2x^2-2) is -sin(2+2x^2+7x^3)+C

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integral of (-21x^2-4x)cos(-7x^3-2x^2-2)

\int (- 21 x^{2} - 4 x) cos (- 7 x^{3} - 2 x^{2} - 2) d x

- sin (2 + 2 x^{2} + 7 x^{3}) + C

Solution steps

\int (- 21 x^{2} - 4 x) cos (- 7 x^{3} - 2 x^{2} - 2) d x

Apply u-substitution

= \int cos (u) d u

Use the common integral:

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

= sin (u)

Substitute back

u = - 7 x^{3} - 2 x^{2} - 2

= sin (- 7 x^{3} - 2 x^{2} - 2)

Use the negative angle identity:

sin (- x) = - sin (x)

= - sin (2 + 2 x^{2} + 7 x^{3})

Add a constant to the solution

= - sin (2 + 2 x^{2} + 7 x^{3}) + C

What is the integral of (-21x^2-4x)cos(-7x^3-2x^2-2) ?
The integral of (-21x^2-4x)cos(-7x^3-2x^2-2) is -sin(2+2x^2+7x^3)+C