What is the integral from x^2 to x^4 of sin(sqrt(t)) ?

The integral from x^2 to x^4 of sin(sqrt(t)) is 2(-x^2cos(x^2)+sin(x^2)+xcos(x)-sin(x))

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integral from x^2 to x^4 of sin(sqrt(t))

\int_{x^{2}}^{x^{4}} sin (t) d t

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

Solution steps

\int_{x^{2}}^{x^{4}} sin (t) d t

Apply u-substitution

: \int_{x^{2}}^{x^{4}} sin (u) \cdot 2 u d u

= \int_{x^{2}}^{x^{4}} sin (u) \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int_{x^{2}}^{x^{4}} sin (u) u d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{x^{2}}^{x^{4}}

= 2 [- u cos (u) - \int - cos (u) d u]_{x^{2}}^{x^{4}}

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

= 2 [- u cos (u) - (- sin (u))]_{x^{2}}^{x^{4}}

Simplify

2 [- u cos (u) - (- sin (u))]_{x^{2}}^{x^{4}} : 2 [- u cos (u) + sin (u)]_{x}^{x^{2}}

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

Compute the boundaries:

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

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

\int_{0}^{1 - y^{2}} x y^{2} d x

\int_{1}^{4} 2 x - 3 d x

\int_{2}^{6} 4 x d x

\int_{0}^{0} ln (z + 2) d z

\int_{0}^{\frac{1}{2}} (sin (π x)) d x

What is the integral from x^2 to x^4 of sin(sqrt(t)) ?
The integral from x^2 to x^4 of sin(sqrt(t)) is 2(-x^2cos(x^2)+sin(x^2)+xcos(x)-sin(x))