What is the integral of 6sin^2(xco)s^2x ?

The integral of 6sin^2(xco)s^2x is (3s^2)/(4c^2o^2)(2c^2o^2x^2-2coxsin(2cox)-cos(2cox))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6sin^2(xco)s^2x

\int 6 sin^{2} (x co) s^{2} x d x

\frac{3 s ^{2}}{4 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

Solution steps

\int 6 sin^{2} (x co) s^{2} x d x

Take the constant out:

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

= 6 s^{2} \cdot \int sin^{2} (co x) x d x

Apply u-substitution

: \int \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

= 6 s^{2} \cdot \int \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

Take the constant out:

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

= 6 s^{2} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

Apply Integration By Parts

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

= 6 s^{2} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u)

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

= 6 s^{2} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u)))

Substitute back

u = co x

= 6 s^{2} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x)))

Simplify

6 s^{2} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x))) : \frac{3 s ^{2}}{4 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

= \frac{3 s ^{2}}{4 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

Add a constant to the solution

= \frac{3 s ^{2}}{4 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

y = 3 x^{3} - x^{2} - 10 x, y = - x^{2} + 2 x

\frac{d}{d x} (\frac{6}{( x - 1 ) ^{2}})

f (x) = 6 cos (x) + 3 sin (2 x)

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 2, y^{^{'}} (0) = 2

\frac{\partial}{\partial x} (1 + 3 x)

What is the integral of 6sin^2(xco)s^2x ?
The integral of 6sin^2(xco)s^2x is (3s^2)/(4c^2o^2)(2c^2o^2x^2-2coxsin(2cox)-cos(2cox))+C