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

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

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

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

1 + cos (1)

Decimal Notation

1.54030 \dots

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int_{1}^{π} sin (v) d v

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

Use the common integral:

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

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

Simplify

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

= [- cos (v)]_{1}^{π}

Compute the boundaries:

1 + cos (1)

= 1 + cos (1)

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\frac{d}{d x} (x^{2} - 5 x - 6)

\frac{d}{d u} ((u - u) (u + u))

\frac{\partial}{\partial x} (\frac{1}{1 - \frac{x}{a} - \frac{y}{b}})

\int \frac{1}{12} x d x

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