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

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

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

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

\frac{1}{2} (- cot (2 x) + 2 tan (2 x) + \frac{1}{3} tan^{3} (2 x)) + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{2 cos ^{4} ( u ) sin ^{2} ( u )} 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{1}{cos ^{4} ( u ) sin ^{2} ( u )} d u

Apply Trigonometric Substitution

= \frac{1}{2} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{v ^{2}} d v

Expand

\frac{( 1 + v ^{2} ) ^{2}}{v ^{2}} : \frac{1}{v ^{2}} + 2 + v^{2}

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

Apply the Sum Rule:

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

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

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

\int 2 d v = 2 v

\int v^{2} d v = \frac{v ^{3}}{3}

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

Substitute back

= \frac{1}{2} (- \frac{1}{tan ( 2 x )} + 2 tan (2 x) + \frac{tan ^{3} ( 2 x )}{3})

Simplify

\frac{1}{2} (- \frac{1}{tan ( 2 x )} + 2 tan (2 x) + \frac{tan ^{3} ( 2 x )}{3}) : \frac{1}{2} (- cot (2 x) + 2 tan (2 x) + \frac{1}{3} tan^{3} (2 x))

= \frac{1}{2} (- cot (2 x) + 2 tan (2 x) + \frac{1}{3} tan^{3} (2 x))

Add a constant to the solution

= \frac{1}{2} (- cot (2 x) + 2 tan (2 x) + \frac{1}{3} tan^{3} (2 x)) + C

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