What is the integral of csc^3(x)cot^2(x) ?

The integral of csc^3(x)cot^2(x) is

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integral of csc^3(x)cot^2(x)

\int csc^{3} (x) cot^{2} (x) d x

- \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x) + C

Solution steps

\int csc^{3} (x) cot^{2} (x) d x

Rewrite using trig identities

= \int \frac{1 - sin ^{2} ( x )}{sin ^{5} ( x )} d x

Expand

\frac{1 - sin ^{2} ( x )}{sin ^{5} ( x )} : \frac{1}{sin ^{5} ( x )} - \frac{1}{sin ^{3} ( x )}

= \int \frac{1}{sin ^{5} ( x )} - \frac{1}{sin ^{3} ( x )} d x

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}{sin ^{5} ( x )} d x - \int \frac{1}{sin ^{3} ( x )} d x

\int \frac{1}{sin ^{5} ( x )} d x = - \frac{cos ( x ) csc ^{4} ( x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (x) csc (x) + \frac{1}{2} ln tan (\frac{x}{2}))

\int \frac{1}{sin ^{3} ( x )} d x = \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

= - \frac{cos ( x ) csc ^{4} ( x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (x) csc (x) + \frac{1}{2} ln tan (\frac{x}{2})) - \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2}))

Simplify

- \frac{cos ( x ) csc ^{4} ( x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (x) csc (x) + \frac{1}{2} ln tan (\frac{x}{2})) - \frac{1}{8} (- cot^{2} (\frac{x}{2}) + 4 ln tan (\frac{x}{2}) + tan^{2} (\frac{x}{2})) : - \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x)

= - \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x)

Add a constant to the solution

= - \frac{3}{8} cot (x) csc (x) - \frac{1}{8} ln tan (\frac{x}{2}) + \frac{1}{8} cot^{2} (\frac{x}{2}) - \frac{1}{8} tan^{2} (\frac{x}{2}) - \frac{1}{4} csc^{4} (x) cos (x) + C

e^{θ}

\int \frac{( e ^{2 x} + 1 )}{e ^{x}} d x

y^{^{'}} = \frac{1}{y}, y (0) = 1

f (x) = 3 x^{2} + 1

\frac{d}{d x} (ln^{3} (2 x + 5))