What is the integral of 4x^4sin(2x) ?

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

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integral of 4x^4sin(2x)

\int 4 x^{4} sin (2 x) d x

- 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x)) + C

Solution steps

\int 4 x^{4} sin (2 x) d x

Take the constant out:

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

= 4 \cdot \int x^{4} sin (2 x) d x

Apply u-substitution

: \int \frac{u ^{4} sin ( u )}{32} d u

= 4 \cdot \int \frac{u ^{4} sin ( u )}{32} d u

Take the constant out:

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

= 4 \cdot \frac{1}{32} \cdot \int u^{4} sin (u) d u

Apply Integration By Parts

: - u^{4} cos (u) - \int - 4 u^{3} cos (u) d u

= 4 \cdot \frac{1}{32} (- u^{4} cos (u) - \int - 4 u^{3} cos (u) d u)

\int - 4 u^{3} cos (u) d u = - 4 (u^{3} sin (u) - 3 (- u^{2} cos (u) + 2 (u sin (u) + cos (u))))

= 4 \cdot \frac{1}{32} (- u^{4} cos (u) - (- 4 (u^{3} sin (u) - 3 (- u^{2} cos (u) + 2 (u sin (u) + cos (u))))))

Substitute back

u = 2 x

= 4 \cdot \frac{1}{32} (- (2 x)^{4} cos (2 x) - (- 4 ((2 x)^{3} sin (2 x) - 3 (- (2 x)^{2} cos (2 x) + 2 (2 x sin (2 x) + cos (2 x))))))

Simplify

4 \cdot \frac{1}{32} (- (2 x)^{4} cos (2 x) - (- 4 ((2 x)^{3} sin (2 x) - 3 (- (2 x)^{2} cos (2 x) + 2 (2 x sin (2 x) + cos (2 x)))))) : - 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))

= - 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))

Add a constant to the solution

= - 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x)) + C

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