What is the integral of sin^4(pix) ?

The integral of sin^4(pix) is 1/pi (-1/4 sin^3(pix)cos(pix)+3/8 (pix-1/2 sin(2pix)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^4(pix)

\int sin^{4} (π x) d x

\frac{1}{π} (- \frac{1}{4} sin^{3} (π x) cos (π x) + \frac{3}{8} (π x - \frac{1}{2} sin (2 π x))) + C

Solution steps

\int sin^{4} (π x) d x

Apply u-substitution

: \int sin^{4} (u) \frac{1}{π} d u

= \int sin^{4} (u) \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{π} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

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

= \frac{1}{π} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = π x

= \frac{1}{π} (- \frac{cos ( π x ) sin ^{3} ( π x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (π x - \frac{1}{2} sin (2 π x)))

Simplify

\frac{1}{π} (- \frac{cos ( π x ) sin ^{3} ( π x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (π x - \frac{1}{2} sin (2 π x))) : \frac{1}{π} (- \frac{1}{4} sin^{3} (π x) cos (π x) + \frac{3}{8} (π x - \frac{1}{2} sin (2 π x)))

= \frac{1}{π} (- \frac{1}{4} sin^{3} (π x) cos (π x) + \frac{3}{8} (π x - \frac{1}{2} sin (2 π x)))

Add a constant to the solution

= \frac{1}{π} (- \frac{1}{4} sin^{3} (π x) cos (π x) + \frac{3}{8} (π x - \frac{1}{2} sin (2 π x))) + C

\int arctan (3 t) d t

\int \frac{9}{sec ( x ) - 1} d x

\int y^{2} sin (n y) d y

\int \frac{e ^{4 v}}{( 1 + 3 e ^{4 v} ) ^{2}} d v

What is the integral of sin^4(pix) ?
The integral of sin^4(pix) is 1/pi (-1/4 sin^3(pix)cos(pix)+3/8 (pix-1/2 sin(2pix)))+C