What is the integral of (sin(2x))/((1+cos(2x))^5) ?

The integral of (sin(2x))/((1+cos(2x))^5) is 1/128 sec^8(x)+C

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integral of (sin(2x))/((1+cos(2x))^5)

\int \frac{sin ( 2 x )}{( 1 + cos ( 2 x ) ) ^{5}} d x

\frac{1}{128} sec^{8} (x) + C

Solution steps

\int \frac{sin ( 2 x )}{( 1 + cos ( 2 x ) ) ^{5}} d x

Apply u-substitution

: \int \frac{sin ( u )}{2 ( 1 + cos ( u ) ) ^{5}} d u

= \int \frac{sin ( u )}{2 ( 1 + cos ( u ) ) ^{5}} 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 ) ) ^{5}} d u

Apply u-substitution

: \int - \frac{1}{v ^{5}} d v

= \frac{1}{2} \cdot \int - \frac{1}{v ^{5}} 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 ^{5}} d v)

Apply the Power Rule

: - \frac{1}{4 v ^{4}}

= \frac{1}{2} (- (- \frac{1}{4 v ^{4}}))

Substitute back

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

Simplify

\frac{1}{2} (- (- \frac{1}{4 ( 1 + cos ( 2 x ) ) ^{4}})) : \frac{1}{128} sec^{8} (x)

= \frac{1}{128} sec^{8} (x)

Add a constant to the solution

= \frac{1}{128} sec^{8} (x) + C

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What is the integral of (sin(2x))/((1+cos(2x))^5) ?
The integral of (sin(2x))/((1+cos(2x))^5) is 1/128 sec^8(x)+C