What is the integral of 7cos(sqrt(5x)) ?

The integral of 7cos(sqrt(5x)) is 14/5 (sqrt(5)sqrt(x)sin(sqrt(5)sqrt(x))+cos(sqrt(5)sqrt(x)))+C

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integral of 7cos(sqrt(5x))

\int 7 cos (5 x) d x

\frac{14}{5} (5 x sin (5 x) + cos (5 x)) + C

Solution steps

\int 7 cos (5 x) d x

Take the constant out:

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

= 7 \cdot \int cos (5 x) d x

Apply u-substitution

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

= 7 \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

= 7 \cdot \frac{1}{5} \cdot \int cos (u) d u

Apply u-substitution

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

= 7 \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

= 7 \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

= 7 \cdot \frac{1}{5} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 7 \cdot \frac{1}{5} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 7 \cdot \frac{1}{5} \cdot 2 (5 x sin (5 x) - (- cos (5 x)))

Simplify

7 \cdot \frac{1}{5} \cdot 2 (5 x sin (5 x) - (- cos (5 x))) : \frac{14}{5} (5 x sin (5 x) + cos (5 x))

= \frac{14}{5} (5 x sin (5 x) + cos (5 x))

Add a constant to the solution

= \frac{14}{5} (5 x sin (5 x) + cos (5 x)) + C

x \to 1 lim (\frac{ln ( x )}{4 x - x ^{2} - 3})

\frac{d}{d x} ((5 - x^{2})^{5})

x \to 0 + lim (x^{- 2})

x \to 2 lim (ln ∣ 2 - x ∣)

\int 5 x^{2} e^{x^{3}} d x

What is the integral of 7cos(sqrt(5x)) ?
The integral of 7cos(sqrt(5x)) is 14/5 (sqrt(5)sqrt(x)sin(sqrt(5)sqrt(x))+cos(sqrt(5)sqrt(x)))+C