What is the integral from 0 to x^2 of sin(sqrt(t)) ?

The integral from 0 to x^2 of sin(sqrt(t)) is 2(sin(x)-xcos(x))

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integral from 0 to x^2 of sin(sqrt(t))

\int_{0}^{x^{2}} sin (t) d t

2 (sin (x) - x cos (x))

Solution steps

\int_{0}^{x^{2}} sin (t) d t

Apply u-substitution

: \int_{0}^{x^{2}} sin (u) \cdot 2 u d u

= \int_{0}^{x^{2}} 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_{0}^{x^{2}} sin (u) u d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{0}^{x^{2}}

= 2 [- u cos (u) - \int - cos (u) d u]_{0}^{x^{2}}

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

= 2 [- u cos (u) - (- sin (u))]_{0}^{x^{2}}

Simplify

2 [- u cos (u) - (- sin (u))]_{0}^{x^{2}} : 2 [- u cos (u) + sin (u)]_{0}^{x}

= 2 [- u cos (u) + sin (u)]_{0}^{x}

Compute the boundaries:

sin (x) - x cos (x)

= 2 (sin (x) - x cos (x))

\frac{5}{x}, \frac{5}{x ^{2}} 1, 7

\int \frac{4 x ^{2} - 3 x + 4}{x ( x ^{2} + 4 )} d x

y = \frac{1}{2} x^{2}, \frac{1}{x ^{2} + 1}

(1 + x) \frac{d y}{d x} - x y = x^{2} + x

\frac{\partial}{\partial y} (3 x^{2} + 2 x y)

What is the integral from 0 to x^2 of sin(sqrt(t)) ?
The integral from 0 to x^2 of sin(sqrt(t)) is 2(sin(x)-xcos(x))