What is the integral from 0 to sqrt(ln(pi) of)2xe^{x^2}cos(e^{x^2}) ?

The integral from 0 to sqrt(ln(pi) of)2xe^{x^2}cos(e^{x^2}) is -sin(1)

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integral from 0 to sqrt(ln(pi) of)2xe^{x^2}cos(e^{x^2})

\int_{0}^{l n (π)} 2 x e^{x^{2}} cos (e^{x^{2}}) d x

- sin (1)

Decimal Notation

- 0.84147 \dots

Solution steps

\int_{0}^{l n (π)} 2 x e^{x^{2}} cos (e^{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_{0}^{l n (π)} x e^{x^{2}} cos (e^{x^{2}}) d x

Apply u-substitution

: \int_{0}^{l n (π)} \frac{e ^{u} cos ( e ^{u} )}{2} d u

= 2 \cdot \int_{0}^{l n (π)} \frac{e ^{u} cos ( e ^{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_{0}^{l n (π)} e^{u} cos (e^{u}) d u

Apply u-substitution

: \int_{1}^{π} cos (v) d v

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{π} cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= 2 \cdot \frac{1}{2} [sin (v)]_{1}^{π}

Simplify

2 \cdot \frac{1}{2} [sin (v)]_{1}^{π} : [sin (v)]_{1}^{π}

= [sin (v)]_{1}^{π}

Compute the boundaries:

- sin (1)

= - sin (1)

x \to - 6 - lim (x^{2} + \frac{2}{x + 6})

\frac{\partial}{\partial x} (1 + 2 x)

9 - x^{2}, x

f (t) = 1 + t^{2}

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

What is the integral from 0 to sqrt(ln(pi) of)2xe^{x^2}cos(e^{x^2}) ?
The integral from 0 to sqrt(ln(pi) of)2xe^{x^2}cos(e^{x^2}) is -sin(1)