What is the integral of (2sin(1/(5x)))/(5x^2) ?

The integral of (2sin(1/(5x)))/(5x^2) is 2cos(1/(5x))+C

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integral of (2sin(1/(5x)))/(5x^2)

\int \frac{2 sin ( \frac{1}{5 x} )}{5 x ^{2}} d x

2 cos (\frac{1}{5 x}) + C

Solution steps

\int \frac{2 sin ( \frac{1}{5 x} )}{5 x ^{2}} d x

Take the constant out:

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

= \frac{2}{5} \cdot \int \frac{sin ( \frac{1}{5 x} )}{x ^{2}} d x

Apply u-substitution

: \int \frac{5 sin ( \frac{1}{u} )}{u ^{2}} d u

= \frac{2}{5} \cdot \int \frac{5 sin ( \frac{1}{u} )}{u ^{2}} d u

Take the constant out:

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

= \frac{2}{5} \cdot 5 \cdot \int \frac{sin ( \frac{1}{u} )}{u ^{2}} d u

Apply u-substitution

: \int - sin (v) d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

= 2 cos (\frac{1}{5 x})

Add a constant to the solution

= 2 cos (\frac{1}{5 x}) + C

y = \frac{64}{16 + x ^{2}}, y = 1

\int 2 sec (4 x) d x

\frac{1}{s + 42} - \frac{1}{( s + 3 ) ^{4}}

\int - π cos (1 π) t d t

\frac{\partial}{\partial x} (- 2 e^{3 x})

What is the integral of (2sin(1/(5x)))/(5x^2) ?
The integral of (2sin(1/(5x)))/(5x^2) is 2cos(1/(5x))+C