What is the integral of 9cos(sqrt(5x)) ?

The integral of 9cos(sqrt(5x)) is 18/5 (sqrt(5)sqrt(x)sin(sqrt(5)sqrt(x))+cos(sqrt(5)sqrt(x)))+C

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integral of 9cos(sqrt(5x))

\int 9 cos (5 x) d x

\frac{18}{5} (5 x sin (5 x) + cos (5 x)) + C

Solution steps

\int 9 cos (5 x) d x

Take the constant out:

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

= 9 \cdot \int cos (5 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{5} d u

= 9 \cdot \int cos (u) \frac{1}{5} d u

Take the constant out:

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

= 9 \cdot \frac{1}{5} \cdot \int cos (u) d u

Apply u-substitution

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

= 9 \cdot \frac{1}{5} \cdot \int cos (v) \cdot 2 v d v

Take the constant out:

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

= 9 \cdot \frac{1}{5} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

: v sin (v) - \int sin (v) d v

= 9 \cdot \frac{1}{5} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 9 \cdot \frac{1}{5} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 9 \cdot \frac{1}{5} \cdot 2 (5 x sin (5 x) - (- cos (5 x)))

Simplify

9 \cdot \frac{1}{5} \cdot 2 (5 x sin (5 x) - (- cos (5 x))) : \frac{18}{5} (5 x sin (5 x) + cos (5 x))

= \frac{18}{5} (5 x sin (5 x) + cos (5 x))

Add a constant to the solution

= \frac{18}{5} (5 x sin (5 x) + cos (5 x)) + C

\frac{d}{d x} (8 e^{6 x})

y^{^{'}} - 5 y = 2 e^{(5 x)}

\int_{0}^{\frac{π}{2}} 5 sin^{2} (2 x) d x

f (x) = 5 - 2 x^{2}, at x = - 3, 13

4 x^{\frac{3}{2}}

What is the integral of 9cos(sqrt(5x)) ?
The integral of 9cos(sqrt(5x)) is 18/5 (sqrt(5)sqrt(x)sin(sqrt(5)sqrt(x))+cos(sqrt(5)sqrt(x)))+C