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

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

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

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

\frac{1}{2} (- \frac{cos ^{3} ( x ^{2} )}{3} + \frac{cos ^{5} ( x ^{2} )}{5}) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Expand

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

= \frac{1}{2} \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}{2} (- \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}{2} (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{2} (- \frac{cos ^{3} ( x ^{2} )}{3} + \frac{cos ^{5} ( x ^{2} )}{5})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{3} ( x ^{2} )}{3} + \frac{cos ^{5} ( x ^{2} )}{5}) + C

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 9

\int x^{2} ln (x) d x

\int_{2}^{7} t^{4} ln (3 t) d t

x^{- 9}

\int e^{r^{2}} r d r

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