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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 7sin^4(x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (x) cos (t))

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

y^{^{''}} - 6 y^{^{'}} = 0

x \to 2 lim (3 x^{3} - 7 x^{2} + 3)

- 9 x^{2} y + \frac{d y}{d x} = - 2 x^{2}

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