What is the integral from 0 to x^2 of 1/x cos(y/x) ?

The integral from 0 to x^2 of 1/x cos(y/x) is sin(x)

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integral from 0 to x^2 of 1/x cos(y/x)

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

sin (x)

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{x} \cdot \int_{0}^{x^{2}} cos (\frac{y}{x}) d y

Apply u-substitution

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

= \frac{1}{x} \cdot \int_{0}^{x} cos (u) x d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{x} x \cdot \int_{0}^{x} cos (u) d u

Use the common integral:

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

= \frac{1}{x} x [sin (u)]_{0}^{x}

Simplify

\frac{1}{x} x [sin (u)]_{0}^{x} : [sin (u)]_{0}^{x}

= [sin (u)]_{0}^{x}

Compute the boundaries:

sin (x)

= sin (x)

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

What is the integral from 0 to x^2 of 1/x cos(y/x) ?
The integral from 0 to x^2 of 1/x cos(y/x) is sin(x)