What is the integral of 1/2 cos(sqrt(x)) ?

The integral of 1/2 cos(sqrt(x)) is sqrt(x)sin(sqrt(x))+cos(sqrt(x))+C

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integral of 1/2 cos(sqrt(x))

\int \frac{1}{2} cos (x) d x

x sin (x) + cos (x) + C

Solution steps

\int \frac{1}{2} cos (x) d x

Take the constant out:

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

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

Apply u-substitution

= \frac{1}{2} \cdot \int 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

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

Apply Integration By Parts

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

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

= \frac{1}{2} \cdot 2 (u sin (u) - (- cos (u)))

Substitute back

u = x

= \frac{1}{2} \cdot 2 (x sin (x) - (- cos (x)))

Simplify

\frac{1}{2} \cdot 2 (x sin (x) - (- cos (x))) : x sin (x) + cos (x)

= x sin (x) + cos (x)

Add a constant to the solution

= x sin (x) + cos (x) + C

\frac{d}{d x} (\frac{3 x}{x ^{3} + 7 x - 5})

f (x) = lo g_{2} (1 - 3 x)

sin (2 x + π)

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{x}, y (0) = 0, y^{^{'}} (0) = 2

\int 3 x^{2} (x^{3} + 1)^{2} d x

What is the integral of 1/2 cos(sqrt(x)) ?
The integral of 1/2 cos(sqrt(x)) is sqrt(x)sin(sqrt(x))+cos(sqrt(x))+C