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

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

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

integral

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Expand

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

= \int \frac{1}{u ^{2}} + 2 + u^{2} d u

Apply the Sum Rule:

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

= \int \frac{1}{u ^{2}} d u + \int 2 d u + \int u^{2} d u

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

\int 2 d u = 2 u

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

= - \frac{1}{u} + 2 u + \frac{u ^{3}}{3}

Substitute back

u = tan (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{x ^{4}}{( x + 1 )}

\frac{( x ^{2} - x - 2 )}{( 1 - x ^{2} )}

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

\frac{x}{( x ^{2} + 9 )}

x (x + 4)^{8}

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