What is the integral of (x^2+2x)cos(x^3+3x^2-47) ?

The integral of (x^2+2x)cos(x^3+3x^2-47) is 1/3 sin(x^3+3x^2-47)+C

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integral of (x^2+2x)cos(x^3+3x^2-47)

\int (x^{2} + 2 x) cos (x^{3} + 3 x^{2} - 47) d x

\frac{1}{3} sin (x^{3} + 3 x^{2} - 47) + C

Solution steps

\int (x^{2} + 2 x) cos (x^{3} + 3 x^{2} - 47) d x

Apply u-substitution

: \int \frac{cos ( u )}{3} d u

= \int \frac{cos ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{3} sin (u)

Substitute back

u = x^{3} + 3 x^{2} - 47

= \frac{1}{3} sin (x^{3} + 3 x^{2} - 47)

Add a constant to the solution

= \frac{1}{3} sin (x^{3} + 3 x^{2} - 47) + C

(7 x + 8) (8 x - 9)

y = \frac{x}{4 + x}

n = 1 \sum \infty \frac{4}{n ( n + 3 ) !}

\int 3 x^{2} y d x

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

What is the integral of (x^2+2x)cos(x^3+3x^2-47) ?
The integral of (x^2+2x)cos(x^3+3x^2-47) is 1/3 sin(x^3+3x^2-47)+C