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

The integral of 1/(sin^5(x)*cos(x)) is -1/4 cot^4(x)-cot^2(x)-ln|cot(x)|+C

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

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

- \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣ + C

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

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

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

(u^{2} + 1) u^{2} + 1 u^{2} + 1 = (u^{2} + 1)^{2}

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

Expand

\frac{( u ^{2} + 1 ) ^{2}}{u} : u^{3} + 2 u + \frac{1}{u}

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

\int u^{3} d u = \frac{u ^{4}}{4}

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

\int \frac{1}{u} d u = ln ∣ u ∣

= - (\frac{u ^{4}}{4} + u^{2} + ln ∣ u ∣)

Substitute back

u = cot (x)

= - (\frac{cot ^{4} ( x )}{4} + cot^{2} (x) + ln ∣ cot (x) ∣)

Simplify

- (\frac{cot ^{4} ( x )}{4} + cot^{2} (x) + ln ∣ cot (x) ∣) : - \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣

= - \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣

Add a constant to the solution

= - \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣ + C

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What is the integral of 1/(sin^5(x)*cos(x)) ?
The integral of 1/(sin^5(x)*cos(x)) is -1/4 cot^4(x)-cot^2(x)-ln|cot(x)|+C