What is the integral of sec^4(x)sqrt(cot^3(x)) ?

The integral of sec^4(x)sqrt(cot^3(x)) is -(-2/3 tan^{3/2}(x)+2sqrt(cot(x)))+C

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integral of sec^4(x)sqrt(cot^3(x))

\int sec^{4} (x) cot^{3} (x) d x

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

Solution steps

\int sec^{4} (x) cot^{3} (x) d x

Apply u-substitution

: \int - \frac{1 + u ^{2}}{u ^{\frac{5}{2}}} d u

= \int - \frac{1 + u ^{2}}{u ^{\frac{5}{2}}} d u

Take the constant out:

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

= - \int \frac{1 + u ^{2}}{u ^{\frac{5}{2}}} d u

Expand

\frac{1 + u ^{2}}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{5}{2}}} + \frac{1}{u ^{\frac{1}{2}}}

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

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

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

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

Substitute back

u = cot (x)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sec^4(x)sqrt(cot^3(x)) ?
The integral of sec^4(x)sqrt(cot^3(x)) is -(-2/3 tan^{3/2}(x)+2sqrt(cot(x)))+C