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

The integral of 2xsin(x^2) is -cos(x^2)+C

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integral of 2xsin(x^2)

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

- cos (x^{2}) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int sin (u) d u = - cos (u)

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

Substitute back

u = x^{2}

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

Simplify

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

= - cos (x^{2})

Add a constant to the solution

= - cos (x^{2}) + C

y^{^{'}} + \frac{y ^{2}}{x} = - \frac{1}{x}

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

(x^{''})^{^{'}}

y^{^{''}} + a \cdot y^{^{'}} = b, y (0) = 0, y^{^{'}} (0) = 0

\frac{d y}{d x} = \frac{( 2 x + sec ^{2} ( x ) )}{2 y}, y (0) = - 6