What is the integral of xcos((x^2+y^2)/2) ?

The integral of xcos((x^2+y^2)/2) is sin(1/2 (y^2+x^2))+C

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integral of xcos((x^2+y^2)/2)

\int x cos (\frac{x ^{2} + y ^{2}}{2}) d x

sin (\frac{1}{2} (y^{2} + x^{2})) + C

Solution steps

\int x cos (\frac{x ^{2} + y ^{2}}{2}) d x

Apply u-substitution

: \int \frac{cos ( \frac{u}{2} )}{2} d u

= \int \frac{cos ( \frac{u}{2} )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (\frac{u}{2}) d u

Apply u-substitution

: \int cos (v) \cdot 2 d v

= \frac{1}{2} \cdot \int cos (v) \cdot 2 d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int cos (v) d v

Use the common integral:

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

= \frac{1}{2} \cdot 2 sin (v)

Substitute back

= \frac{1}{2} \cdot 2 sin (\frac{y ^{2} + x ^{2}}{2})

Simplify

\frac{1}{2} \cdot 2 sin (\frac{y ^{2} + x ^{2}}{2}) : sin (\frac{1}{2} (y^{2} + x^{2}))

= sin (\frac{1}{2} (y^{2} + x^{2}))

Add a constant to the solution

= sin (\frac{1}{2} (y^{2} + x^{2})) + C

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\int \frac{1}{3} x e^{\frac{x}{6}} d x

What is the integral of xcos((x^2+y^2)/2) ?
The integral of xcos((x^2+y^2)/2) is sin(1/2 (y^2+x^2))+C