What is the integral of 2cos(sqrt(10x)) ?

The integral of 2cos(sqrt(10x)) is 2/5 (sqrt(10)sqrt(x)sin(sqrt(10)sqrt(x))+cos(sqrt(10)sqrt(x)))+C

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integral of 2cos(sqrt(10x))

\int 2 cos (10 x) d x

\frac{2}{5} (10 x sin (10 x) + cos (10 x)) + C

Solution steps

\int 2 cos (10 x) d x

Take the constant out:

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

= 2 \cdot \int cos (10 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{10} d u

= 2 \cdot \int cos (u) \frac{1}{10} d u

Take the constant out:

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

= 2 \cdot \frac{1}{10} \cdot \int cos (u) d u

Apply u-substitution

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

= 2 \cdot \frac{1}{10} \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

= 2 \cdot \frac{1}{10} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

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

= 2 \cdot \frac{1}{10} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 2 \cdot \frac{1}{10} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= 2 \cdot \frac{1}{10} \cdot 2 (10 x sin (10 x) - (- cos (10 x)))

Simplify

2 \cdot \frac{1}{10} \cdot 2 (10 x sin (10 x) - (- cos (10 x))) : \frac{2}{5} (10 x sin (10 x) + cos (10 x))

= \frac{2}{5} (10 x sin (10 x) + cos (10 x))

Add a constant to the solution

= \frac{2}{5} (10 x sin (10 x) + cos (10 x)) + C

\frac{d}{d x} ((x^{2} + sin (x))^{3})

\int_{- 1}^{1} x^{3} - x d x

1 e^{4 t}

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

\frac{6 x}{x + 7}, at x = 1

What is the integral of 2cos(sqrt(10x)) ?
The integral of 2cos(sqrt(10x)) is 2/5 (sqrt(10)sqrt(x)sin(sqrt(10)sqrt(x))+cos(sqrt(10)sqrt(x)))+C