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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u sin ^{3} ( u ^{\frac{3}{2}} ) cos ^{2} ( u ^{\frac{3}{2}} )}{2 t} 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} ( t ^{3} )}{3} + \frac{cos ^{5} ( t ^{3} )}{5})

Add a constant to the solution

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

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

tan^{2} (x)

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

\int \frac{x}{( 1 - x ) ^{3}} d x

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