What is the integral of (sin^4(x))/((1+cos(x))^2) ?

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

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integral of (sin^4(x))/((1+cos(x))^2)

integral

\frac{sin ^{4} ( x )}{( 1 + cos ( x ) ) ^{2}}

\frac{3}{2} x + \frac{1}{4} sin (2 x) - 2 sin (x) + C

Solution steps

\int \frac{sin ^{4} ( x )}{( 1 + cos ( x ) ) ^{2}} d x

Multiply by the conjugate of

1 + cos (x) : \frac{sin ^{2} ( x )}{1 - cos ( x )}

= \int \frac{sin ^{4} ( x )}{( \frac{s i n ^{2} ( x )}{1 - c o s ( x )} ) ^{2}} d x

Simplify

\frac{sin ^{4} ( x )}{( \frac{s i n ^{2} ( x )}{1 - c o s ( x )} ) ^{2}} : (1 - cos (x))^{2}

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

Expand

(1 - cos (x))^{2} : 1 - 2 cos (x) + cos^{2} (x)

= \int 1 - 2 cos (x) + cos^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x - \int 2 cos (x) d x + \int cos^{2} (x) d x

\int 1 d x = x

\int 2 cos (x) d x = 2 sin (x)

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

= x - 2 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

x - 2 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{3}{2} x + \frac{1}{4} sin (2 x) - 2 sin (x)

= \frac{3}{2} x + \frac{1}{4} sin (2 x) - 2 sin (x)

Add a constant to the solution

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

in t e g r a l \frac{1}{( 2 x ^{3} )}

in t e g r a l sec^{2} (2 x - 1)

in t e g r a l \frac{1}{( k + e ^{- t} )}

in t e g r a l e^{\frac{y ^{2}}{2}}

in t e g r a l x (x^{2})

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