What is the integral of 1/(cot^3(x)) ?

The integral of 1/(cot^3(x)) is 1/2 sec^2(x)+ln|cos(x)|+C

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integral of 1/(cot^3(x))

integral

\frac{1}{cot ^{3} ( x )}

\frac{1}{2} sec^{2} (x) + ln ∣ cos (x) ∣ + C

Solution steps

\int \frac{1}{cot ^{3} ( x )} d x

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ^{3} ( x )} d x

Apply u-substitution

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

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

Expand

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

= - (- \frac{1}{2 u ^{2}}) + ln ∣ u ∣

Substitute back

u = cos (x)

= - (- \frac{1}{2 cos ^{2} ( x )}) + ln ∣ cos (x) ∣

Simplify

- (- \frac{1}{2 cos ^{2} ( x )}) + ln ∣ cos (x) ∣ : \frac{1}{2} sec^{2} (x) + ln ∣ cos (x) ∣

= \frac{1}{2} sec^{2} (x) + ln ∣ cos (x) ∣

Add a constant to the solution

= \frac{1}{2} sec^{2} (x) + ln ∣ cos (x) ∣ + C

\frac{x}{( x + b ) ^{3}}

y = 2 e^{4 x - 3} 6

lo g_{10} (x^{2})

\frac{( x ^{5} + 1 )}{( x - 1 )}

\frac{e ^{x^{2}}}{x}

What is the integral of 1/(cot^3(x)) ?
The integral of 1/(cot^3(x)) is 1/2 sec^2(x)+ln|cos(x)|+C