What is the integral of x^2sin^3(x^3)cos^2(x^3) ?

The integral of x^2sin^3(x^3)cos^2(x^3) is 1/3 (-(cos^3(x^3))/3+(cos^5(x^3))/5)+C

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integral of x^2sin^3(x^3)cos^2(x^3)

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

\frac{1}{3} (- \frac{cos ^{3} ( x ^{3} )}{3} + \frac{cos ^{5} ( x ^{3} )}{5}) + C

Solution steps

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

Apply u-substitution

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

= \int \frac{sin ^{3} ( u ) cos ^{2} ( 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 sin^{3} (u) cos^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- v^{2} (1 - v^{2}) : - v^{2} + v^{4}

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

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

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

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

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

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{3} ( x ^{3} )}{3} + \frac{cos ^{5} ( x ^{3} )}{5}) + C

1

\int \frac{2}{( x + 1 ) ( x + 3 )} d x

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

What is the integral of x^2sin^3(x^3)cos^2(x^3) ?
The integral of x^2sin^3(x^3)cos^2(x^3) is 1/3 (-(cos^3(x^3))/3+(cos^5(x^3))/5)+C