What is the integral of 7cos(sqrt(4x)) ?

The integral of 7cos(sqrt(4x)) is 7sqrt(x)sin(2sqrt(x))+7/2 cos(2sqrt(x))+C

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integral of 7cos(sqrt(4x))

\int 7 cos (4 x) d x

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

Solution steps

\int 7 cos (4 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 (4 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{4} d u

= 7 \cdot \int cos (u) \frac{1}{4} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int cos (u) d u

Apply u-substitution

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

= 7 \cdot \frac{1}{4} \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}{4} \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}{4} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 7 \cdot \frac{1}{4} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 7 \cdot \frac{1}{4} \cdot 2 (4 x sin (4 x) - (- cos (4 x)))

Simplify

7 \cdot \frac{1}{4} \cdot 2 (4 x sin (4 x) - (- cos (4 x))) : 7 x sin (2 x) + \frac{7}{2} cos (2 x)

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

Add a constant to the solution

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

f (x) = 3 x^{4} - 7 ln (x)

\int sin^{3} (2 x) cot (2 x) d x

\frac{\partial}{\partial y} (y^{2} + x)

\frac{1}{s} - \frac{s}{s ^{2} + 1}

What is the integral of 7cos(sqrt(4x)) ?
The integral of 7cos(sqrt(4x)) is 7sqrt(x)sin(2sqrt(x))+7/2 cos(2sqrt(x))+C