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

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