What is the integral of 17sin^2(xco)s^2x ?

The integral of 17sin^2(xco)s^2x is (17s^2)/(8c^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

integral of 17sin^2(xco)s^2x

\int 17 sin^{2} (x co) s^{2} x d x

\frac{17 s ^{2}}{8 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 17 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

= 17 s^{2} \cdot \int sin^{2} (co x) x d x

Apply u-substitution

= 17 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

= 17 s^{2} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

Apply Integration By Parts

= 17 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))

= 17 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

= 17 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

17 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{17 s ^{2}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

= \frac{17 s ^{2}}{8 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{17 s ^{2}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

What is the integral of 17sin^2(xco)s^2x ?
The integral of 17sin^2(xco)s^2x is (17s^2)/(8c^2o^2)(2c^2o^2x^2-2coxsin(2cox)-cos(2cox))+C