Question Number 111868 by mathdave last updated on 05/Sep/20
$${simplify} \\ $$$$\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{5}} {x}\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$
Answered by Her_Majesty last updated on 05/Sep/20
$$\int\frac{\left(\mathrm{1}−\mathrm{5}{sin}^{\mathrm{2}} {x}\right){cos}^{\mathrm{3}} {x}}{{cos}^{\mathrm{8}} {xsin}^{\mathrm{2}} {x}}{dx} \\ $$$$\int\frac{{v}'{u}}{{v}^{\mathrm{2}} }=−\frac{{u}}{{v}}+\int\frac{{u}'}{{v}} \\ $$$${v}={cos}^{\mathrm{4}} {xsinx}\:\Rightarrow\:{v}'=\left(\mathrm{1}−\mathrm{5}{sin}^{\mathrm{2}} {x}\right){cos}^{\mathrm{3}} {x} \\ $$$${u}'=\mathrm{0}\:\Rightarrow\:{u}=\mathrm{1} \\ $$$$\int\frac{\left(\mathrm{1}−\mathrm{5}{sin}^{\mathrm{2}} {x}\right){cos}^{\mathrm{3}} {x}}{{cos}^{\mathrm{8}} {xsin}^{\mathrm{2}} {x}}{dx}=−\frac{\mathrm{1}}{{cos}^{\mathrm{4}} {xsinx}}+{C} \\ $$
Commented by MJS_new last updated on 05/Sep/20
great!
Commented by Her_Majesty last updated on 05/Sep/20
$${thank}\:{you} \\ $$
Commented by mathdave last updated on 05/Sep/20
$${that}\:{is}\:{not}\:{the}\:{answer} \\ $$
Commented by Her_Majesty last updated on 05/Sep/20
$${answer}\:{as}\:{you}\:{wish},\:{I}\:{just}\:{solved}\:{the}\:{integral} \\ $$
Commented by MJS_new last updated on 05/Sep/20
Sir mathdave please tell us, what is the answer?
Answered by mathdave last updated on 06/Sep/20
![solution to ∫((1−5sin^2 x)/(cos^5 xsin^2 x))dx I=∫(1/(cos^5 xsin^2 x))dx−∫((5sin^2 x)/(cos^5 xsin^2 x))dx I=∫sec^5 xcosec^2 x−5∫sec^5 x=A−5B.....(1) A=∫sec^5 xcosec^2 xdx (but cosec^2 x=1+cot^2 x) A=∫sec^5 xdx+∫sec^5 xcot^2 xdx=B+∫sec^3 xcosec^2 xdx A=∫sec^3 x(1+cot^2 x)dx+B=∫secxcosec^2 xdx+∫sec^3 xdx+B let C=∫sec^3 xdx A=∫secx(1+cot^2 x)dx+C+B=∫cotxcosecxdx+∫secxdx+B+C but ∫secxdx=ln(secx+tanx) A=∫cotxcosecxdx+ln(secx+tanx)+B+C but note ∫sec^n xdx=((sec^(n−2) xtanx)/(n−1))+((n−2)/(n−1))∫sec^(n−2) dx note sinh^(−1) (tanx)=ln(tanx+secx) when used sinh^(−1) x=ln(x+(√(1+x^2 ))) we have C=∫sec^3 xdx=(1/2)[sinh^(−1) (tanx)+tanxsecx] A=∫cotxcosecx+ln(tanx+secx)+(1/2)[sinh^(−1) (tanx)+tanxsecx]+B but but ∫cosec^n (ax)cot(ax)dx=−((cosec^n (ax))/(na))+k ∵∫cotxcosecxdx=−cosecx+k A=−cosecx+ln(tanx+secx)+(1/2)[sinh^(−1) (tanx)+tanxsecx]+B ∵I=−cosecx+sinh^(−1) (tanx)+(1/2)[sinh^(−1) (tanx)+tanxsecx]+B−5B Let E=−4B=−4∫sec^5 xdx when using the general formular ∫sec^n xdx=((sec^(n−2) tanx)/(n−1))+((n−2)/(n−1))∫sec^(n−2) dx E=−[sec^3 xtanx+(3/2)(sinh^(−1) (tanx)+tanxsecx)]+k hence I=−cosecx+sinh^(−1) (tanx)+(1/2)[sinh^(−1) (tanx)+secxtanx]−[sec^3 xtanx+(3/2)(sinh^(−1) (tanx)+secxtanx)]+k ∵∫((1−5sin^2 x)/(cos^5 xsin^2 x))dx=sinh^(−1) (tanx)−sec^3 xtanx−cosecx−[sinh^(−1) (tanx)+secxtanx]+k by mathdave(06/09/2020)](https://www.tinkutara.com/question/Q112090.png)
$${solution}\:{to}\:\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{5}} {x}\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$$${I}=\int\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{5}} {x}\mathrm{sin}^{\mathrm{2}} {x}}{dx}−\int\frac{\mathrm{5sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{5}} {x}\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$$${I}=\int\mathrm{sec}^{\mathrm{5}} {x}\mathrm{cosec}^{\mathrm{2}} {x}−\mathrm{5}\int\mathrm{sec}^{\mathrm{5}} {x}={A}−\mathrm{5}{B}…..\left(\mathrm{1}\right) \\ $$$${A}=\int\mathrm{sec}^{\mathrm{5}} {x}\mathrm{cosec}^{\mathrm{2}} {xdx}\:\:\:\:\left({but}\:\mathrm{cosec}^{\mathrm{2}} {x}=\mathrm{1}+\mathrm{cot}^{\mathrm{2}} {x}\right) \\ $$$${A}=\int\mathrm{sec}^{\mathrm{5}} {xdx}+\int\mathrm{sec}^{\mathrm{5}} {x}\mathrm{cot}^{\mathrm{2}} {xdx}={B}+\int\mathrm{sec}^{\mathrm{3}} {x}\mathrm{cosec}^{\mathrm{2}} {xdx} \\ $$$${A}=\int\mathrm{sec}^{\mathrm{3}} {x}\left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} {x}\right){dx}+{B}=\int\mathrm{sec}{x}\mathrm{cosec}^{\mathrm{2}} {xdx}+\int\mathrm{sec}^{\mathrm{3}} {xdx}+{B} \\ $$$${let}\:{C}=\int\mathrm{sec}^{\mathrm{3}} {xdx} \\ $$$${A}=\int\mathrm{sec}{x}\left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} {x}\right){dx}+{C}+{B}=\int\mathrm{cot}{x}\mathrm{cosec}{xdx}+\int\mathrm{sec}{xdx}+{B}+{C} \\ $$$${but}\:\int\mathrm{sec}{xdx}=\mathrm{ln}\left(\mathrm{sec}{x}+\mathrm{tan}{x}\right) \\ $$$${A}=\int\mathrm{cot}{x}\mathrm{cosec}{xdx}+\mathrm{ln}\left(\mathrm{sec}{x}+\mathrm{tan}{x}\right)+{B}+{C} \\ $$$${but}\:{note}\:\int\mathrm{sec}^{{n}} {xdx}=\frac{\mathrm{sec}^{{n}−\mathrm{2}} {x}\mathrm{tan}{x}}{{n}−\mathrm{1}}+\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\int\mathrm{sec}^{{n}−\mathrm{2}} {dx} \\ $$$${note}\:\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)=\mathrm{ln}\left(\mathrm{tan}{x}+\mathrm{sec}{x}\right)\:{when}\:{used}\:\mathrm{sinh}^{−\mathrm{1}} {x}=\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$${we}\:{have}\:{C}=\int\mathrm{sec}^{\mathrm{3}} {xdx}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{tan}{x}\mathrm{sec}{x}\right] \\ $$$${A}=\int\mathrm{cot}{x}\mathrm{cosec}{x}+\mathrm{ln}\left(\mathrm{tan}{x}+\mathrm{sec}{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{tan}{x}\mathrm{sec}{x}\right]+{B} \\ $$$${but}\:{but}\:\int\mathrm{cosec}^{{n}} \left({ax}\right)\mathrm{cot}\left({ax}\right){dx}=−\frac{\mathrm{cosec}^{{n}} \left({ax}\right)}{{na}}+{k} \\ $$$$\because\int\mathrm{cot}{x}\mathrm{cosec}{xdx}=−\mathrm{cosec}{x}+{k} \\ $$$${A}=−\mathrm{cosec}{x}+\mathrm{ln}\left(\mathrm{tan}{x}+\mathrm{sec}{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{tan}{x}\mathrm{sec}{x}\right]+{B} \\ $$$$\because{I}=−\mathrm{cosec}{x}+\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{tan}{x}\mathrm{sec}{x}\right]+{B}−\mathrm{5}{B} \\ $$$${Let}\:\:{E}=−\mathrm{4}{B}=−\mathrm{4}\int\mathrm{sec}^{\mathrm{5}} {xdx} \\ $$$${when}\:{using}\:{the}\:{general}\:{formular} \\ $$$$\int\mathrm{sec}^{{n}} {xdx}=\frac{\mathrm{sec}^{{n}−\mathrm{2}} \mathrm{tan}{x}}{{n}−\mathrm{1}}+\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\int\mathrm{sec}^{{n}−\mathrm{2}} {dx} \\ $$$${E}=−\left[\mathrm{sec}^{\mathrm{3}} {x}\mathrm{tan}{x}+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{tan}{x}\mathrm{sec}{x}\right)\right]+{k} \\ $$$${hence} \\ $$$${I}=−\mathrm{cosec}{x}+\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{sec}{x}\mathrm{tan}{x}\right]−\left[\mathrm{sec}^{\mathrm{3}} {x}\mathrm{tan}{x}+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{sec}{x}\mathrm{tan}{x}\right)\right]+{k} \\ $$$$\because\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{5}} {x}\mathrm{sin}^{\mathrm{2}} {x}}{dx}=\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)−\mathrm{sec}^{\mathrm{3}} {x}\mathrm{tan}{x}−\mathrm{cosec}{x}−\left[\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{sec}{x}\mathrm{tan}{x}\right]+{k} \\ $$$${by}\:{mathdave}\left(\mathrm{06}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by MJS_new last updated on 06/Sep/20
![ok but (d/dx)[−(1/(cos^4 x sin x))]=−((−4cos^3 x sin^2 x +cos^5 x)/(cos^8 x sin^2 x))= =((cos^2 x −4sin^2 x)/(cos^5 x sin^2 x))=((1−5sin^2 x)/(cos^5 x sin^2 x))](https://www.tinkutara.com/question/Q112112.png)
$$\mathrm{ok}\:\mathrm{but} \\ $$$$\frac{{d}}{{dx}}\left[−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}}\right]=−\frac{−\mathrm{4cos}\:^{\mathrm{3}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{cos}^{\mathrm{5}} \:{x}}{\mathrm{cos}^{\mathrm{8}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}= \\ $$$$=\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{4sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{5}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}=\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{5}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$
Commented by MJS_new last updated on 06/Sep/20
![you have not finished your solution sinh^(−1) (tanx)−sec^3 xtanx−cosecx−[sinh^(−1) (tanx)+secxtanx]= =−sec^3 x tan x −cosec x −sec x tan x= =−((sin x)/(cos^4 x))−(1/(sin x))−((sin x)/(cos^2 x))=−((sin^2 x +cos^4 x +cos^2 x sin^2 x)/(cos^4 x sin x))= =−(1/(cos^4 x sin x)) so your path leads to the same result but it′s an unnecessary detour compared to the solution of “Her Majesty”](https://www.tinkutara.com/question/Q112118.png)
$$\mathrm{you}\:\mathrm{have}\:\mathrm{not}\:\mathrm{finished}\:\mathrm{your}\:\mathrm{solution} \\ $$$$\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)−\mathrm{sec}^{\mathrm{3}} {x}\mathrm{tan}{x}−\mathrm{cosec}{x}−\left[\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{tan}{x}\right)+\mathrm{sec}{x}\mathrm{tan}{x}\right]= \\ $$$$=−\mathrm{sec}^{\mathrm{3}} \:{x}\:\mathrm{tan}\:{x}\:−\mathrm{cosec}\:{x}\:−\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}= \\ $$$$=−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{4}} \:{x}}−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=−\frac{\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}} \\ $$$$\mathrm{so}\:\mathrm{your}\:\mathrm{path}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{same}\:\mathrm{result}\:\mathrm{but}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{an}\:\mathrm{unnecessary}\:\mathrm{detour}\:\mathrm{compared}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{solution}\:\mathrm{of}\:“\mathrm{Her}\:\mathrm{Majesty}'' \\ $$
Commented by Her_Majesty last updated on 06/Sep/20
$${mathdave},\:{you}\:{proved}\:{that}\:{you}\:{can}\:{solve} \\ $$$${an}\:{integral}\:{within}\:\mathrm{26}\:{lines}\:{that}\:{I}\:{solved} \\ $$$${within}\:\mathrm{5}\:{lines}\:{only}\:{to}\:{get}\:{the}\:{same}\:{answer} \\ $$$${you}\:{claimed}\:{to}\:{be}\:{wrong}.\:{how}\:{great}\:{thou}\:{art}! \\ $$
Commented by mathdave last updated on 06/Sep/20
$${who}\:{told}\:{u}\:{i}\:{moved}\:{in}\:{an}\:{unnecessary} \\ $$$${detour}\:{hey}\:{dont}\:{ever}\:{derogate}\:{my} \\ $$$${working}\: \\ $$
Commented by Her_Majesty last updated on 06/Sep/20
$${it}\:{seems}\:{a}\:{lot}\:{to}\:{me}\:{you}\:{simply}\:{don}'{t}\:{know} \\ $$$${the}\:{method}\:“{solving}\:{by}\:{parts}'' \\ $$$$ \\ $$$$\left({uv}\right)'={u}'{v}+{v}'{u}\:\Leftrightarrow\:{uv}=\int{u}'{v}+\int{v}'{u}\:\Leftrightarrow \\ $$$$\int{u}'{v}={uv}−\int{v}'{u} \\ $$$$ \\ $$$$\left(\frac{{u}}{{v}}\right)'=\frac{{u}'{v}−{v}'{u}}{{v}^{\mathrm{2}} }=\frac{{u}'}{{v}}−\frac{{v}'{u}}{{v}^{\mathrm{2}} }\:\Leftrightarrow\:\frac{{u}}{{v}}=\int\frac{{u}'}{{v}}−\int\frac{{v}'{u}}{{v}^{\mathrm{2}} }\:\Leftrightarrow \\ $$$$\begin{cases}{\int\frac{{u}'}{{v}}=\frac{{u}}{{v}}+\int\frac{{v}'{u}}{{v}^{\mathrm{2}} }}\\{\int\frac{{v}'{u}}{{v}^{\mathrm{2}} }=−\frac{{u}}{{v}}+\int\frac{{u}'}{{v}}}\end{cases} \\ $$$${the}\:{last}\:{one}\:{I}\:{used} \\ $$
Commented by mathdave last updated on 06/Sep/20
$${smirks},{u}\:{thought}\:{am}\:{a}\:{neophyte}\:{in} \\ $$$${mathematics}\: \\ $$
Commented by Her_Majesty last updated on 06/Sep/20
$${well}\:{you}\:{needed}\:\mathrm{420\%}\:{more}\:{space}\:{for}\:{your} \\ $$$${solution},\:{this}\:{is}\:{undeniable} \\ $$
Commented by mathdave last updated on 06/Sep/20
$${ur}\:{working}\:{doesnt}\:{intrique}\:{me}\:{so}\:{i}\:{dont} \\ $$$${give}\:{damn}\:{to}\:{it}\:{madam}\: \\ $$
Commented by MJS_new last updated on 06/Sep/20
What made you such a vitriolic young man?
Commented by MJS_new last updated on 06/Sep/20
...she gave the right answer, you stated it was wrong but your answer is the same as hers. Does this mean, your answer is wrong, too? You assume she's copied her answer from Google just because you don't like (or you didn't know) the method she used?!
Commented by mathdave last updated on 07/Sep/20
$${i}\:{cant}\:{used}\:{this}\:{filthy}\:{or}\:{scanty}\:{way}\:{to} \\ $$$${express}\:{my}\:{work}\:{bcos}\:{that}\:{wasnt}\:{the} \\ $$$${way}\:{i}\:{was}\:{taught}\:{to}\:{work}\:{mathematics}\:{problem} \\ $$$${from}\:{the}\:{university}\:{i}\:{graduated}\:{from} \\ $$
Commented by mathdave last updated on 07/Sep/20
$${her}\:{way}\:{was}\:{cheap}\:{for}\:{me},{y}\:{wont}\:{i}\:{not} \\ $$$${kwn}\:{this}\:{her}\:{way}\:{wen}\:{am}\:{nt}\:{a}\:{toddler} \\ $$$${in}\:{mathematics} \\ $$
Commented by MJS_new last updated on 07/Sep/20
![In my opinion the art of mathematics is to find the solution at least possible cost. ∫((1−5sin^2 x)/(cos^5 x sin^2 x))dx can also be solved by using Weierstrass′ Substitution but it′s also lengthy: i∫((1−5sin^2 x)/(cos^5 x sin^2 x))dx= [t=tan (x/2) → dx=((2dt)/(t^2 +1))] =−(1/2)∫(((t^2 +1)^4 (t^2 −4t−1)(t^2 +4t−1))/(t^2 (t−1)^5 (t+1)^5 ))dt= =∫((4/((t−1)^5 ))+(6/((t−1)^4 ))+(6/((t−1)^3 ))+(2/((t−1)^2 ))−(4/((t+1)^5 ))+(6/((t+1)^4 ))−(6/((t+1)^3 ))+(2/((t+1)^2 ))+(1/(2t^2 ))−(1/2))dt= =−(1/((t−1)^4 ))−(2/((t−1)^3 ))−(3/((t−1)^2 ))−(2/(t−1))+(1/((t+1)^4 ))−(2/((t+1)^3 ))+(3/((t+1)^2 ))−(2/(t+1))−(1/(2t))−(t/2)= =−(((t^2 +1)^5 )/(2t(t−1)^4 (t+1)^4 ))= [t=(s/c); s=sin (x/2) ∧ c=cos (x/2)] =−(((s^2 +c^2 )^5 )/(2cs(s^2 −c^2 )^4 ))=−(1/(sin x cos^4 x))+C or, much shorter: ∫((1−5sin^2 x)/(cos^5 x sin^2 x))dx= [t=sin x → dx=(dt/(cos x))] =∫((1−5t^2 )/(t^2 (1−t^2 )^3 ))dt= =∫((1/(2(t−1)^3 ))−(1/(2(t−1)^2 ))−(1/(2(t+1)^3 ))−(1/(2(t+1)^2 ))+(1/t^2 ))dt= =−(1/(4(t−1)^2 ))+(1/(2(t−1)))+(1/(4(t+1)^2 ))+(1/(2(t+1)))−(1/t)= =−(1/(t(t^2 −1)^2 ))=−(1/(sin x cos^4 x))+C btw. also I have not seen the possibility to use Integration by Parts](https://www.tinkutara.com/question/Q112304.png)
$$\mathrm{In}\:\mathrm{my}\:\mathrm{opinion}\:\mathrm{the}\:\mathrm{art}\:\mathrm{of}\:\mathrm{mathematics}\:\mathrm{is}\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{at}\:\mathrm{least}\:\mathrm{possible}\:\mathrm{cost}. \\ $$$$\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{5}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{using} \\ $$$$\mathrm{Weierstrass}'\:\mathrm{Substitution}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{lengthy}: \\ $$$$\mathrm{i}\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{5}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} \left({t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{4}{t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)^{\mathrm{5}} \left({t}+\mathrm{1}\right)^{\mathrm{5}} }{dt}= \\ $$$$=\int\left(\frac{\mathrm{4}}{\left({t}−\mathrm{1}\right)^{\mathrm{5}} }+\frac{\mathrm{6}}{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{6}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{4}}{\left({t}+\mathrm{1}\right)^{\mathrm{5}} }+\frac{\mathrm{6}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} }−\frac{\mathrm{6}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\right){dt}= \\ $$$$=−\frac{\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }−\frac{\mathrm{2}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{3}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{{t}−\mathrm{1}}+\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} }−\frac{\mathrm{2}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{3}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{{t}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{t}}−\frac{{t}}{\mathrm{2}}= \\ $$$$=−\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{5}} }{\mathrm{2}{t}\left({t}−\mathrm{1}\right)^{\mathrm{4}} \left({t}+\mathrm{1}\right)^{\mathrm{4}} }= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{s}}{{c}};\:{s}=\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:\wedge\:{c}=\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right] \\ $$$$=−\frac{\left({s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{5}} }{\mathrm{2}{cs}\left({s}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{4}} }=−\frac{\mathrm{1}}{\mathrm{sin}\:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}}+{C} \\ $$$$ \\ $$$$\mathrm{or},\:\mathrm{much}\:\mathrm{shorter}: \\ $$$$\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{5}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cos}\:{x}}\right] \\ $$$$=\int\frac{\mathrm{1}−\mathrm{5}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt}= \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{t}}= \\ $$$$=−\frac{\mathrm{1}}{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{sin}\:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}}+{C} \\ $$$$\mathrm{btw}.\:\mathrm{also}\:\mathrm{I}\:\mathrm{have}\:\mathrm{not}\:\mathrm{seen}\:\mathrm{the}\:\mathrm{possibility}\:\mathrm{to} \\ $$$$\mathrm{use}\:\mathrm{Integration}\:\mathrm{by}\:\mathrm{Parts} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$