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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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

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

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

\int_{0}^{2} \frac{3 x + 1}{x} d x

\int_{0}^{1} \frac{2 x + 3}{x ^{2} - x + 1} d x

\int_{- 3}^{2} 4 x^{3} - 2 x^{2} + x + 5 d x

\int_{0.4}^{5.2} k x^{2} d x

\int_{0}^{3} s^{3} 81 - s^{4} d s

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