What is the integral of 3xcos^3(x^2+2) ?

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

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integral of 3xcos^3(x^2+2)

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

\frac{3}{4} sin (2 x^{2}) - \frac{1}{4} sin^{3} (2 x^{2}) + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int x cos^{3} (x^{2} + 2) d x

Rewrite using trig identities

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

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= 3 \cdot \int x cos^{3} (2 x^{2}) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int 1 - v^{2} d v

= 3 \cdot \frac{1}{4} \cdot \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{4} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= 3 \cdot \frac{1}{4} (v - \frac{v ^{3}}{3})

Substitute back

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

Simplify

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

= \frac{3}{4} sin (2 x^{2}) - \frac{1}{4} sin^{3} (2 x^{2})

Add a constant to the solution

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

f (x) = x^{2}, at m (x) = - \frac{1}{2}

\frac{d}{d x} (r^{2} - 8 x^{2})

\int x (x + 4)^{- \frac{3}{2}} d x

n = 1 \sum \infty (- 1)^{n} n^{- 3}

\int \frac{1}{x ^{2} x ^{2} - 81} d x

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