What is the integral from-1 to x of sin(t^2)2t ?

The integral from-1 to x of sin(t^2)2t is -cos(x^2)+cos(1)

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integral from-1 to x of sin(t^2)2t

\int_{- 1}^{x} sin (t^{2}) 2 t d t

- cos (x^{2}) + cos (1)

Solution steps

\int_{- 1}^{x} sin (t^{2}) \cdot 2 t d t

Take the constant out:

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

= 2 \cdot \int_{- 1}^{x} sin (t^{2}) t d t

Apply u-substitution

: \int_{1}^{x^{2}} \frac{sin ( u )}{2} d u

= 2 \cdot \int_{1}^{x^{2}} \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_{1}^{x^{2}} sin (u) d u

Use the common integral:

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

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

Simplify

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

= [- cos (u)]_{1}^{x^{2}}

Compute the boundaries:

- cos (x^{2}) + cos (1)

= - cos (x^{2}) + cos (1)

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\int_{2}^{5} 3 x^{2} d x

\int_{π}^{2 π} (x - 2 sin (x)) d x

What is the integral from-1 to x of sin(t^2)2t ?
The integral from-1 to x of sin(t^2)2t is -cos(x^2)+cos(1)