What is the integral of xe^{x^2}cos(x^2) ?

The integral of xe^{x^2}cos(x^2) is 1/4 (e^{x^2}sin(x^2)+e^{x^2}cos(x^2))+C

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integral of xe^{x^2}cos(x^2)

\int x e^{x^{2}} cos (x^{2}) d x

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

Solution steps

\int x e^{x^{2}} cos (x^{2}) d x

Apply u-substitution

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

= \int \frac{e ^{u} cos ( 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 e^{u} cos (u) d u

Apply Integration By Parts

: e^{u} sin (u) - \int e^{u} sin (u) d u

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

Apply Integration By Parts

: - e^{u} cos (u) - \int - e^{u} cos (u) d u

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{u} cos (u) d u

= \frac{1}{2} (\frac{e ^{u} sin ( u )}{2} + \frac{e ^{u} cos ( u )}{2})

Substitute back

u = x^{2}

= \frac{1}{2} (\frac{e ^{x^{2}} sin ( x ^{2} )}{2} + \frac{e ^{x^{2}} cos ( x ^{2} )}{2})

Simplify

\frac{1}{2} (\frac{e ^{x^{2}} sin ( x ^{2} )}{2} + \frac{e ^{x^{2}} cos ( x ^{2} )}{2}) : \frac{1}{4} (e^{x^{2}} sin (x^{2}) + e^{x^{2}} cos (x^{2}))

= \frac{1}{4} (e^{x^{2}} sin (x^{2}) + e^{x^{2}} cos (x^{2}))

Add a constant to the solution

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

\frac{d}{d x} (\frac{1 - 2 x}{3 x ^{2}})

x \to 1 lim (\frac{2}{x ^{3} - x})

\frac{d}{d x} ((\frac{x + 5}{x ^{2} + 2})^{2})

\frac{d ^{2}}{d x ^{2}} (e^{\frac{6}{x}})

x \to \infty lim (- 3 x^{3} + x)

What is the integral of xe^{x^2}cos(x^2) ?
The integral of xe^{x^2}cos(x^2) is 1/4 (e^{x^2}sin(x^2)+e^{x^2}cos(x^2))+C