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

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

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

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

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

Solution steps

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

Use the common integral:

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

= [- cos (x)]_{1}^{x^{2}}

Compute the boundaries:

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

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

\int_{0}^{1} (2 x)^{2} d x

\int_{1}^{8} \frac{14 x}{x ^{2} + 6} d x

\int_{0}^{5} 50 - 2 x^{2} d x

\int_{0}^{1} 2^{x} e^{x} d x

\int_{- 1}^{0} 2 x - 1 d x

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