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

The integral of (sin(2x))/((1+cos(2x))^3) is 1/16 sec^4(x)+C

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integral of (sin(2x))/((1+cos(2x))^3)

\int \frac{sin ( 2 x )}{( 1 + cos ( 2 x ) ) ^{3}} d x

\frac{1}{16} sec^{4} (x) + C

Solution steps

\int \frac{sin ( 2 x )}{( 1 + cos ( 2 x ) ) ^{3}} d x

Apply u-substitution

: \int \frac{sin ( u )}{2 ( 1 + cos ( u ) ) ^{3}} d u

= \int \frac{sin ( u )}{2 ( 1 + cos ( u ) ) ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sin ( u )}{( 1 + cos ( u ) ) ^{3}} d u

Apply u-substitution

: \int - \frac{1}{v ^{3}} d v

= \frac{1}{2} \cdot \int - \frac{1}{v ^{3}} d v

Take the constant out:

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

= \frac{1}{2} (- \int \frac{1}{v ^{3}} d v)

Apply the Power Rule

: - \frac{1}{2 v ^{2}}

= \frac{1}{2} (- (- \frac{1}{2 v ^{2}}))

Substitute back

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

Simplify

\frac{1}{2} (- (- \frac{1}{2 ( 1 + cos ( 2 x ) ) ^{2}})) : \frac{1}{16} sec^{4} (x)

= \frac{1}{16} sec^{4} (x)

Add a constant to the solution

= \frac{1}{16} sec^{4} (x) + C

f (x) = \frac{x ^{5}}{5} + \frac{1}{12 x ^{3}}

y = sin (2 x) + sin (2) (2 x)

x \to - \frac{π}{2} lim (2 tan (x))

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

\frac{d}{d t} (- 2 sin (t))

What is the integral of (sin(2x))/((1+cos(2x))^3) ?
The integral of (sin(2x))/((1+cos(2x))^3) is 1/16 sec^4(x)+C