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

The integral of x^4sin(2x) is 1/4 (-2x^4cos(2x)+4x^3sin(2x)-3(-2x^2cos(2x)+2xsin(2x)+cos(2x)))+C

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integral of x^4sin(2x)

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

\frac{1}{4} (- 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 x^{4} sin (2 x) d x

Apply u-substitution

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

= \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

= \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

= \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))))

= \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

= \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

\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)))))) : \frac{1}{4} (- 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)))

= \frac{1}{4} (- 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

= \frac{1}{4} (- 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

f (x) = arcsin (6 x^{4})

n = 1 \sum \infty n e^{n}

\int_{2}^{5} t^{3} ln (5 t) d t

\frac{d}{d x} ((x - 1) (x - 2) (2 - x^{2}))

\frac{3}{6 t + 3}

What is the integral of x^4sin(2x) ?
The integral of x^4sin(2x) is 1/4 (-2x^4cos(2x)+4x^3sin(2x)-3(-2x^2cos(2x)+2xsin(2x)+cos(2x)))+C