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

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

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integral of xsin^2(x^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

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} \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

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

= \frac{1}{2} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{e ^{x}}{e ^{2 x} + 4 e ^{x} + 3} d x

x \to 0 lim (\frac{60}{x ^{4}})

\frac{d}{d x} (\frac{sec ( x )}{x})

x^{c o s (x)}

- 5 t^{2} y^{^{''}} + t y^{^{'}} + 8 y = 0

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