How do you evaluate $\int {cot}^{5} (2 x) d x$ $int cot^5(2x) dx$ ?

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How do you evaluate $\int {cot}^{5} (2 x) d x$ $int cot^5(2x) dx$ ?

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Answer 1 · 2018-04-19T19:33:22

$I = \frac{1}{8} (4 ln (sin (2 x)) - {csc}^{4} (2 x) + 4 {csc}^{2} (2 x)) + C$ $I=1/8(4ln(sin(2x))-csc^4(2x)+4csc^2(2x))+C$

Explanation:

We want to integrate

$I = \int {cot}^{5} (2 x) d x$ $I=intcot^5(2x)dx$

Make a substitution $u = 2 x \Rightarrow d u = 2 d x$ $color(red)(u=2x=>du=2dx$

$I = \frac{1}{2} \int {cot}^{5} (u) d u$ $I=1/2intcot^5(u)du$

$I = \frac{1}{2} \int \frac{{({cos}^{2} (u))}^{2} cos (u)}{{sin}^{5} (u)} d u$ $color(white)(I)=1/2int((cos^2(u))^2cos(u))/sin^5(u)du$

$I = \frac{1}{2} \int \frac{{(1 - {sin}^{2} (u))}^{2} cos (u)}{{sin}^{5} (u)} d u$ $color(white)(I)=1/2int((1-sin^2(u))^2cos(u))/sin^5(u)du$

Make a substitution $s = sin (u) \Rightarrow d s = cos (u) d u$ $color(red)(s=sin(u)=>ds=cos(u)du$

$I = \frac{1}{2} \int \frac{{(1 - s^{2})}^{2}}{s^{5}} d s$ $I=1/2int(1-s^2)^2/s^5ds$

$I = \frac{1}{2} \int \frac{1}{s} + \frac{1}{s^{5}} - 2 \frac{1}{s^{3}} d s$ $color(white)(I)=1/2int1/s+1/s^5-2 1/s^3ds$

$I = \frac{1}{2} (ln (s) - \frac{1}{4 s^{4}} + \frac{1}{s^{2}}) + C$ $color(white)(I)=1/2(ln(s)-1/(4s^4)+1/s^2)+C$

$I = \frac{1}{8} (4 ln (s) - \frac{1}{s^{4}} + \frac{4}{s^{2}}) + C$ $color(white)(I)=1/8(4ln(s)-1/s^4+4/s^2)+C$

Substitute back $s = sin (u)$ $color(red)(s=sin(u)$ and $u = 2 x$ $color(red)(u=2x$

$I = \frac{1}{8} (4 ln (sin (2 x)) - {csc}^{4} (2 x) + 4 {csc}^{2} (2 x)) + C$ $I=1/8(4ln(sin(2x))-csc^4(2x)+4csc^2(2x))+C$

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