What is the integral of 6x^2sin(x^3-5) ?

The integral of 6x^2sin(x^3-5) is 2(x^3sin(x^3-5)+x^3sin(5-x^3)-cos(5-x^3))+C

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integral of 6x^2sin(x^3-5)

\int 6 x^{2} sin (x^{3} - 5) d x

2 (x^{3} sin (x^{3} - 5) + x^{3} sin (5 - x^{3}) - cos (5 - x^{3})) + C

Solution steps

\int 6 x^{2} sin (x^{3} - 5) d x

Take the constant out:

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

= 6 \cdot \int x^{2} sin (x^{3} - 5) d x

Apply u-substitution

: \int \frac{u sin ( u ^{\frac{3}{2}} - 5 )}{2 x} d u

= 6 \cdot \int \frac{sin ( u - 5 )}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int sin (u - 5) d u

Apply Integration By Parts

: sin (u - 5) u - \int cos (u - 5) u d u

= 6 \cdot \frac{1}{3} (sin (u - 5) u - \int cos (u - 5) u d u)

\int cos (u - 5) u d u = - u sin (5 - u) + cos (5 - u)

= 6 \cdot \frac{1}{3} (sin (u - 5) u - (- u sin (5 - u) + cos (5 - u)))

Substitute back

u = x^{3}

= 6 \cdot \frac{1}{3} (sin (x^{3} - 5) x^{3} - (- x^{3} sin (5 - x^{3}) + cos (5 - x^{3})))

Simplify

6 \cdot \frac{1}{3} (sin (x^{3} - 5) x^{3} - (- x^{3} sin (5 - x^{3}) + cos (5 - x^{3}))) : 2 (x^{3} sin (x^{3} - 5) + x^{3} sin (5 - x^{3}) - cos (5 - x^{3}))

= 2 (x^{3} sin (x^{3} - 5) + x^{3} sin (5 - x^{3}) - cos (5 - x^{3}))

Add a constant to the solution

= 2 (x^{3} sin (x^{3} - 5) + x^{3} sin (5 - x^{3}) - cos (5 - x^{3})) + C

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What is the integral of 6x^2sin(x^3-5) ?
The integral of 6x^2sin(x^3-5) is 2(x^3sin(x^3-5)+x^3sin(5-x^3)-cos(5-x^3))+C