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

The integral from 0 to x^2 of sin(t) is -cos(x^2)+1

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

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

- cos (x^{2}) + 1

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- cos (x^{2}) + 1

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

\frac{d}{d x} (\frac{x ^{\frac{3}{2}}}{3} - x^{\frac{1}{2}})

\frac{7}{r ^{3}}

f (t) = 8 t sin (π t)

\frac{d}{d x} (x^{2 n})

\frac{d}{d x} (\frac{1}{9 - x ^{2}})

What is the integral from 0 to x^2 of sin(t) ?
The integral from 0 to x^2 of sin(t) is -cos(x^2)+1