What is the integral from 1 to x^3 of sin(t) ?

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

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integral from 1 to x^3 of sin(t)

\int_{1}^{x^{3}} sin (t) d t

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

Solution steps

\int_{1}^{x^{3}} sin (t) d t

Use the common integral:

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

= [- cos (t)]_{1}^{x^{3}}

Compute the boundaries:

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

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

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

\int_{1}^{2} \frac{y}{2} d y

\int_{0}^{2} 6 e^{3 x} d x

\int_{- 2}^{0} x^{2} + 2 x d x

\int_{2}^{4} x^{3} - 6 x^{2} + 8 x d x

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