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

The integral of 1/((sin(x))^5(cos(x))^5) is

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

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

6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}} + C

Solution steps

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

Simplify

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

Apply u-substitution

: \int \frac{1}{u ^{5} ( 1 - u ^{2} ) ^{3}} d u

= \int \frac{1}{u ^{5} ( 1 - u ^{2} ) ^{3}} d u

Take the partial fraction of

\frac{1}{u ^{5} ( 1 - u ^{2} ) ^{3}} : \frac{6}{u} + \frac{3}{u ^{3}} + \frac{1}{u ^{5}} - \frac{3}{u + 1} - \frac{13}{16 ( u + 1 ) ^{2}} - \frac{1}{8 ( u + 1 ) ^{3}} - \frac{3}{u - 1} + \frac{13}{16 ( u - 1 ) ^{2}} - \frac{1}{8 ( u - 1 ) ^{3}}

= \int \frac{6}{u} + \frac{3}{u ^{3}} + \frac{1}{u ^{5}} - \frac{3}{u + 1} - \frac{13}{16 ( u + 1 ) ^{2}} - \frac{1}{8 ( u + 1 ) ^{3}} - \frac{3}{u - 1} + \frac{13}{16 ( u - 1 ) ^{2}} - \frac{1}{8 ( u - 1 ) ^{3}} 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{6}{u} d u + \int \frac{3}{u ^{3}} d u + \int \frac{1}{u ^{5}} d u - \int \frac{3}{u + 1} d u - \int \frac{13}{16 ( u + 1 ) ^{2}} d u - \int \frac{1}{8 ( u + 1 ) ^{3}} d u - \int \frac{3}{u - 1} d u + \int \frac{13}{16 ( u - 1 ) ^{2}} d u - \int \frac{1}{8 ( u - 1 ) ^{3}} d u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

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

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

\int \frac{3}{u + 1} d u = 3 ln ∣ u + 1 ∣

\int \frac{13}{16 ( u + 1 ) ^{2}} d u = - \frac{13}{16 ( u + 1 )}

\int \frac{1}{8 ( u + 1 ) ^{3}} d u = - \frac{1}{16 ( u + 1 ) ^{2}}

\int \frac{3}{u - 1} d u = 3 ln ∣ u - 1 ∣

\int \frac{13}{16 ( u - 1 ) ^{2}} d u = - \frac{13}{16 ( u - 1 )}

\int \frac{1}{8 ( u - 1 ) ^{3}} d u = - \frac{1}{16 ( u - 1 ) ^{2}}

= 6 ln ∣ u ∣ - \frac{3}{2 u ^{2}} - \frac{1}{4 u ^{4}} - 3 ln ∣ u + 1 ∣ - (- \frac{13}{16 ( u + 1 )}) - (- \frac{1}{16 ( u + 1 ) ^{2}}) - 3 ln ∣ u - 1 ∣ - \frac{13}{16 ( u - 1 )} - (- \frac{1}{16 ( u - 1 ) ^{2}})

Substitute back

u = sin (x)

= 6 ln ∣ sin (x) ∣ - \frac{3}{2 sin ^{2} ( x )} - \frac{1}{4 sin ^{4} ( x )} - 3 ln ∣ sin (x) + 1 ∣ - (- \frac{13}{16 ( sin ( x ) + 1 )}) - (- \frac{1}{16 ( sin ( x ) + 1 ) ^{2}}) - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} - (- \frac{1}{16 ( sin ( x ) - 1 ) ^{2}})

Simplify

6 ln ∣ sin (x) ∣ - \frac{3}{2 sin ^{2} ( x )} - \frac{1}{4 sin ^{4} ( x )} - 3 ln ∣ sin (x) + 1 ∣ - (- \frac{13}{16 ( sin ( x ) + 1 )}) - (- \frac{1}{16 ( sin ( x ) + 1 ) ^{2}}) - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} - (- \frac{1}{16 ( sin ( x ) - 1 ) ^{2}}) : 6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}}

= 6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}}

Add a constant to the solution

= 6 ln ∣ sin (x) ∣ - \frac{3}{2} csc^{2} (x) - \frac{1}{4} csc^{4} (x) - 3 ln ∣ sin (x) + 1 ∣ + \frac{13}{16 ( sin ( x ) + 1 )} + \frac{1}{16 ( sin ( x ) + 1 ) ^{2}} - 3 ln ∣ sin (x) - 1 ∣ - \frac{13}{16 ( sin ( x ) - 1 )} + \frac{1}{16 ( sin ( x ) - 1 ) ^{2}} + C

\frac{1}{19 s ^{2} + 5 s + 19990}

\frac{\partial}{\partial s} (arctan (rs))

y^{^{''}} + 6 y^{^{'}} + 45 y = 0

y = (x + 9) (3 x + 4)

\int (3 x - 2 3 x) d x

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