What is the integral of sin^4(7x) ?

The integral of sin^4(7x) is 1/7 (-1/4 sin^3(7x)cos(7x)+3/8 (7x-1/2 sin(14x)))+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(7x)

\int sin^{4} (7 x) d x

\frac{1}{7} (- \frac{1}{4} sin^{3} (7 x) cos (7 x) + \frac{3}{8} (7 x - \frac{1}{2} sin (14 x))) + C

Solution steps

\int sin^{4} (7 x) d x

Apply u-substitution

: \int sin^{4} (u) \frac{1}{7} d u

= \int sin^{4} (u) \frac{1}{7} d u

Take the constant out:

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

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

Apply Integral Reduction

= \frac{1}{7} (- \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}{7} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 7 x

= \frac{1}{7} (- \frac{cos ( 7 x ) sin ^{3} ( 7 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (7 x - \frac{1}{2} sin (2 \cdot 7 x)))

Simplify

\frac{1}{7} (- \frac{cos ( 7 x ) sin ^{3} ( 7 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (7 x - \frac{1}{2} sin (2 \cdot 7 x))) : \frac{1}{7} (- \frac{1}{4} sin^{3} (7 x) cos (7 x) + \frac{3}{8} (7 x - \frac{1}{2} sin (14 x)))

= \frac{1}{7} (- \frac{1}{4} sin^{3} (7 x) cos (7 x) + \frac{3}{8} (7 x - \frac{1}{2} sin (14 x)))

Add a constant to the solution

= \frac{1}{7} (- \frac{1}{4} sin^{3} (7 x) cos (7 x) + \frac{3}{8} (7 x - \frac{1}{2} sin (14 x))) + C

\frac{1}{( s - 5 ) ^{2}}

\frac{d}{d x} (x 1 + x^{2})

y^{^{''}} + 4 y^{^{'}} + 13 y = 0

\frac{d ^{2}}{d x ^{2}} (e^{2 x})

f (x) = 2 x - 4, 3, 6

What is the integral of sin^4(7x) ?
The integral of sin^4(7x) is 1/7 (-1/4 sin^3(7x)cos(7x)+3/8 (7x-1/2 sin(14x)))+C