What is the integral from 0 to x^6 of cos(sqrt(t)) ?

The integral from 0 to x^6 of cos(sqrt(t)) is 2(x^3sin(x^3)+cos(x^3)-1)

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integral from 0 to x^6 of cos(sqrt(t))

\int_{0}^{x^{6}} cos (t) d t

2 (x^{3} sin (x^{3}) + cos (x^{3}) - 1)

Solution steps

\int_{0}^{x^{6}} cos (t) d t

Apply u-substitution

= \int_{0}^{x^{6}} cos (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^{6}} cos (u) u d u

Apply Integration By Parts

= 2 [u sin (u) - \int sin (u) d u]_{0}^{x^{6}}

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

= 2 [u sin (u) - (- cos (u))]_{0}^{x^{6}}

Simplify

2 [u sin (u) - (- cos (u))]_{0}^{x^{6}} : 2 [u sin (u) + cos (u)]_{0}^{x^{3}}

= 2 [u sin (u) + cos (u)]_{0}^{x^{3}}

Compute the boundaries:

x^{3} sin (x^{3}) + cos (x^{3}) - 1

= 2 (x^{3} sin (x^{3}) + cos (x^{3}) - 1)

What is the integral from 0 to x^6 of cos(sqrt(t)) ?
The integral from 0 to x^6 of cos(sqrt(t)) is 2(x^3sin(x^3)+cos(x^3)-1)