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

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

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

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

- \frac{1}{2} csc^{4} (x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{8 sin ( 2 x )}{( - cos ( 2 x ) + 1 ) ^{3}} d x

Take the constant out:

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

= 8 \cdot \int \frac{sin ( 2 x )}{( - cos ( 2 x ) + 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= 8 \cdot \int \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{2} (- \frac{1}{2 u ^{2}})

Substitute back

u = - cos (2 x) + 1

= 8 \cdot \frac{1}{2} (- \frac{1}{2 ( - cos ( 2 x ) + 1 ) ^{2}})

Simplify

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

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

Add a constant to the solution

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

\frac{- 3}{( s + 3 ) ^{2}}

17 e^{x}

\frac{\partial}{\partial t} (e^{- t})

\int cos (3 x) sin (x) d x

(9 x^{6} + 4 x^{3})^{4}

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