What is the integral of-2/(pin)cos((pin)/2 x) ?

The integral of-2/(pin)cos((pin)/2 x) is -4/(pi^2n^2)sin((pin)/2 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-2/(pin)cos((pin)/2 x)

\int - \frac{2}{πn} cos (\frac{πn}{2} x) d x

- \frac{4}{π ^{2} n ^{2}} sin (\frac{πn}{2} x) + C

Solution steps

\int - \frac{2}{πn} cos (\frac{πn}{2} x) d x

Take the constant out:

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

= - \frac{2}{πn} \cdot \int cos (\frac{πn}{2} x) d x

Apply u-substitution

: \int \frac{2 cos ( u )}{πn} d u

= - \frac{2}{πn} \cdot \int \frac{2 cos ( u )}{πn} d u

Take the constant out:

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

= - \frac{2}{πn} \cdot \frac{2}{πn} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= - \frac{2}{πn} \cdot \frac{2}{πn} sin (u)

Substitute back

u = \frac{πn}{2} x

= - \frac{2}{πn} \cdot \frac{2}{πn} sin (\frac{πn}{2} x)

Simplify

- \frac{2}{πn} \cdot \frac{2}{πn} sin (\frac{πn}{2} x) : - \frac{4}{π ^{2} n ^{2}} sin (\frac{πn}{2} x)

= - \frac{4}{π ^{2} n ^{2}} sin (\frac{πn}{2} x)

Add a constant to the solution

= - \frac{4}{π ^{2} n ^{2}} sin (\frac{πn}{2} x) + C

\frac{d y}{d x} = \frac{7 x ^{2}}{5 y ^{2}}

f (x) = (\frac{( x + 1 )}{( x - 1 )})^{2}

\int 7 tan^{2} (x) sec (x) d x

x \to 0 lim (\frac{2}{1 + 2 ^{\frac{1}{x}}})

\int \frac{1}{x ^{2} + 16 x} d x

What is the integral of-2/(pin)cos((pin)/2 x) ?
The integral of-2/(pin)cos((pin)/2 x) is -4/(pi^2n^2)sin((pin)/2 x)+C