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

The integral of cos(sqrt(7x)) is 2/7 (sqrt(7)sqrt(x)sin(sqrt(7)sqrt(x))+cos(sqrt(7)sqrt(x)))+C

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integral of cos(sqrt(7x))

\int cos (7 x) d x

\frac{2}{7} (7 x sin (7 x) + cos (7 x)) + C

Solution steps

\int cos (7 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{7} d u

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{7} \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

= \frac{1}{7} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

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

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

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

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

Substitute back

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

Simplify

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

= \frac{2}{7} (7 x sin (7 x) + cos (7 x))

Add a constant to the solution

= \frac{2}{7} (7 x sin (7 x) + cos (7 x)) + C

x^{2} - 4 x + 1, 3 x - 5, 1, 4

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

\int 5 ln (x^{2} + 5) d x

f (x) = (1 + 2 x)^{12}, at x = 0

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0

What is the integral of cos(sqrt(7x)) ?
The integral of cos(sqrt(7x)) is 2/7 (sqrt(7)sqrt(x)sin(sqrt(7)sqrt(x))+cos(sqrt(7)sqrt(x)))+C