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

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

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integral of 2xsin^2(x^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \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

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

Rewrite using trig identities

= 2 \cdot \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

= 2 \cdot \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

= 2 \cdot \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)

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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