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

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

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

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

- sin (1) (\frac{π}{2} C (\frac{2}{π} x^{2}) - \frac{π}{2} C (\frac{2}{π} \cdot 1)) + cos (1) (\frac{π}{2} § (\frac{2}{π} x^{2}) - \frac{π}{2} § (\frac{2}{π} \cdot 1))

Solution steps

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

Rewrite using trig identities

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= - sin (1) (\frac{π}{2} C (\frac{2}{π} x^{2}) - \frac{π}{2} C (\frac{2}{π} \cdot 1)) + cos (1) (\frac{π}{2} § (\frac{2}{π} x^{2}) - \frac{π}{2} § (\frac{2}{π} \cdot 1))

\int_{1}^{3} \frac{2 - 7 x}{x ^{3}} d x

\int_{- 2}^{\infty} e^{- x} d x

\int_{0}^{2 π} sin (3 x) d x

\int_{0}^{\infty} 2 x d x

\int_{- 1}^{1} cos (π x) d x

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