Integrate csc^3 2x dx ?

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Integrate csc^3 2x dx ?

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Answer 1 · 2018-04-04T14:12:38

$I = \frac{1}{16} ({tan}^{2} (x) + 4 ln (tan (x)) - \frac{1}{{tan}^{2} (x)}) + C$ $I=1/16(tan^2(x)+4ln(tan(x))-1/(tan^2(x)))+C$

Explanation:

We want to solve

$I = \int {csc}^{3} (2 x) d x$ $I=intcsc^3(2x)dx$

Make a substitution $u = 2 x \Rightarrow d u = 2 d x$ $u=2x=>du=2dx$

$I = \frac{1}{2} \int {csc}^{3} (u) d u$ $I=1/2intcsc^3(u)du$

Use tangent half-angle substitution $s = tan (\frac{u}{2})$ $s=tan(u/2)$ ,
then $csc (u) = \frac{1 + s^{2}}{2 s}$ $csc(u)=(1+s^2)/(2s)$ and $d u = \frac{2}{1 + s^{2}} d s$ $du=2/(1+s^2)ds$

$I = \frac{1}{2} \int {(\frac{1 + s^{2}}{2 s})}^{3} \frac{2}{1 + s^{2}} d s$ $I=1/2int((1+s^2)/(2s))^3 2/(1+s^2)ds$

$I = \int \frac{{(1 + s^{2})}^{2}}{8 s^{3}} d s$ $color(white)(I)=int(1+s^2)^2/(8s^3)ds$

$I = \frac{1}{8} \int \frac{s^{4} + 2 s^{2} + 1}{s^{3}} d s$ $color(white)(I)=1/8int(s^4+2s^2+1)/(s^3)ds$

$I = \frac{1}{8} \int s + 2 s^{- 1} + s^{- 3} d s$ $color(white)(I)=1/8ints+2s^-1+s^-3ds$

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

$I = \frac{1}{16} (s^{2} + 4 ln (s) - \frac{1}{s^{2}}) + C$ $color(white)(I)=1/16(s^2+4ln(s)-1/(s^2))+C$

Substitute back $s = tan (\frac{u}{2})$ $s=tan(u/2)$ and $u = 2 x$ $u=2x$

$I = \frac{1}{16} ({tan}^{2} (x) + 4 ln (tan (x)) - \frac{1}{{tan}^{2} (x)}) + C$ $I=1/16(tan^2(x)+4ln(tan(x))-1/(tan^2(x)))+C$

Answer 2 · 2018-04-04T14:26:55

$I = - \frac{1}{4} [csc (2 x) cot (2 x) + ln | csc (2 x) + cot (2 x) |] + c$ $I=-1/4[csc(2x)cot(2x)+ln|csc(2x)+cot(2x)|]+c$

Explanation:

Here,

$I = \int {csc}^{3} 2 x d x$ $I=intcsc^3 2x dx$

$= \int (csc 2 x) ({csc}^{2} (2 x)) d x$ $=int(csc2x)(csc^2(2x))dx$

$= \int (\sqrt{{cot}^{2} (2 x) + 1}) {csc}^{2} (2 x) d x$ $=int(sqrt(cot^2(2x)+1))csc^2(2x)dx$

Let,

$cot (2 x) = u \Rightarrow - {csc}^{2} (2 x) \cdot 2 d x = d u$ $color(blue)(cot(2x)=u)=>-csc^2(2x)*2dx=du$

$\Rightarrow {csc}^{2} (2 x) d x = - \frac{1}{2} d u$ $=>csc^2(2x)dx=-1/2du$

$I = \int \sqrt{u^{2} + 1} (- \frac{1}{2}) d u$ $I=intsqrt(u^2+1)(-1/2)du$

We know that,

$\int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} ln ∣ ∣ x + \sqrt{x^{2} + a^{2}} ∣ ∣ + c$ $color(red)(intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c$

$I = - \frac{1}{2} \int \sqrt{u^{2} + 1^{2}} d u$ $I=-1/2intsqrt(u^2+1^2)du$

$= - \frac{1}{2} [\frac{u}{2} \sqrt{u^{2} + 1} + \frac{1}{2} ln ∣ ∣ u + \sqrt{u^{2} + 1} ∣ ∣] + c$ $=-1/2[u/2sqrt(u^2+1)+1/2ln|u+sqrt(u^2+1)|]+c$

substituting back $u = cot 2 x$ $color(blue)(u=cot2x$

$= - \frac{1}{2} [\frac{cot (2 x)}{2} \sqrt{{cot}^{2} (2 x) + 1} + \frac{1}{2} ln ∣ ∣ ∣ cot (2 x) + \sqrt{{cot}^{2} (2 x) + 1} ∣ ∣ ∣] + c$ $=-1/2[cot(2x)/2sqrt(cot^2(2x)+1)+1/2ln|cot(2x)+sqrt(cot^2(2x)+1 )|]+c$

$= - \frac{1}{2} [\frac{cot (2 x)}{2} csc (2 x) + \frac{1}{2} ln | cot (2 x) + csc (2 x) |] + c$ $=-1/2[cot(2x)/2csc(2x)+1/2ln|cot(2x)+csc(2x)|]+c$

$= - \frac{1}{4} [csc (2 x) cot (2 x) + ln | csc (2 x) + cot (2 x) |] + c$ $=-1/4[csc(2x)cot(2x)+ln|csc(2x)+cot(2x)|]+c$

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Integrate csc^3 2x dx ?

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