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Question Number 65950 by ajfour last updated on 06/Aug/19

∫_0 ^( π/2) tan^3 xdx = ?

$$\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{tan}\:^{\mathrm{3}} {xdx}\:=\:? \\ $$

Answered by Tupac Shakur last updated on 06/Aug/19

=∫_0 ^(π/2) ((sin^3 (x))/(cos^3 (x)))dx=∫_0 ^(π/2) ((sin(x){1−cos^2 (x)})/(cos^3 (x)))dx let u=cos(x)==>du=−sin(x)dx ∫_0 ^(π/2) ((sin(x){1−cos^2 (x)})/(cos^3 (x)))dx=∫_0 ^1 ((1−u^2 )/u^3 )du=lim_(t→0) ∫_t ^1 (1/u^3 )−(1/u^ )du=limt→0{−(3/2)+(1/(2t^2 ))+ln(t)}=+∞

$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{3}} \left({x}\right)}{{cos}^{\mathrm{3}} \left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({x}\right)\left\{\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right)\right\}}{{cos}^{\mathrm{3}} \left({x}\right)}{dx} \\ $$ $${let}\:{u}={cos}\left({x}\right)==>{du}=−{sin}\left({x}\right){dx} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({x}\right)\left\{\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right)\right\}}{{cos}^{\mathrm{3}} \left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{u}^{\mathrm{2}} }{{u}^{\mathrm{3}} }{du}={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\int_{{t}} ^{\mathrm{1}} \frac{\mathrm{1}}{{u}^{\mathrm{3}} }−\frac{\mathrm{1}}{{u}^{} }{du}={limt}\rightarrow\mathrm{0}\left\{−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }+{ln}\left({t}\right)\right\}=+\infty \\ $$ $$ \\ $$ $$ \\ $$

Answered by Tanmay chaudhury last updated on 06/Aug/19

∫tan^2 x.tanxdx ∫(sec^2 x−1)tanxdx ∫tanx.d(tanx)−∫tanxdx ((tan^2 x)/2)−lnsecx+c when we put x=(π/2) expression→∞ so some doubt ...

$$\int{tan}^{\mathrm{2}} {x}.{tanxdx} \\ $$ $$\int\left({sec}^{\mathrm{2}} {x}−\mathrm{1}\right){tanxdx} \\ $$ $$\int{tanx}.{d}\left({tanx}\right)−\int{tanxdx} \\ $$ $$\frac{{tan}^{\mathrm{2}} {x}}{\mathrm{2}}−{lnsecx}+{c} \\ $$ $${when}\:{we}\:{put}\:{x}=\frac{\pi}{\mathrm{2}}\:{expression}\rightarrow\infty \\ $$ $${so}\:{some}\:{doubt}\:... \\ $$ $$ \\ $$

Answered by ajfour last updated on 06/Aug/19

∫_0 ^( π/2) ((tan^3 xsec^2 xdx)/((1+tan^2 x))) let tan x=t ∫_0 ^( ∞) ((t^3 dt)/(1+t^2 )) = ∫_0 ^∞ ((t(t^2 +1−1)dt)/(1+t^2 )) =∫_0 ^( c→∞) [(tdt)−(1/2)(((2tdt)/(1+t^2 )))] = lim_(b→∞) [(b/2)−(1/2)ln (1+b)] f(b)=lim_(b→∞) ((ln (1+b))/b)=lim_(b→∞) (1/(1+b)) →0 ⇒ Integral →∞ (but answer in book R.S. Agarwal for class XII is a finite one !)

$$\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\frac{\mathrm{tan}\:^{\mathrm{3}} {x}\mathrm{sec}\:^{\mathrm{2}} {xdx}}{\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)} \\ $$ $${let}\:\:\mathrm{tan}\:{x}={t} \\ $$ $$\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\mathrm{3}} {dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}\left({t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}\right){dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$ $$=\int_{\mathrm{0}} ^{\:{c}\rightarrow\infty} \left[\left({tdt}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)\right] \\ $$ $$\:=\:\underset{{b}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{b}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+{b}\right)\right] \\ $$ $${f}\left({b}\right)=\underset{{b}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{b}\right)}{{b}}=\underset{{b}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}+{b}}\:\rightarrow\mathrm{0} \\ $$ $$\Rightarrow\:{Integral}\:\rightarrow\infty \\ $$ $$\left({but}\:{answer}\:{in}\:{book}\:{R}.{S}.\:{Agarwal}\right. \\ $$ $$\left.{for}\:{class}\:{XII}\:{is}\:{a}\:{finite}\:{one}\:!\right) \\ $$ $$ \\ $$

Commented byMJS last updated on 06/Aug/19

post his answer please...

$$\mathrm{post}\:\mathrm{his}\:\mathrm{answer}\:\mathrm{please}... \\ $$

Answered by MJS last updated on 06/Aug/19

∫tan^n x dx=(1/(n−1))tan^(n−1) x −∫tan^(n−2) x dx ∫tan^3 x dx=(1/2)tan^2 x −∫tan x dx= =(1/2)tan^2 x +ln cos x +C ∫_0 ^(π/2) tan^n x dx=+∞ for n∈N^★ what′s obvious because tan^n x ≥1 for (π/4)≤x<(π/2) and n∈N^★

$$\int\mathrm{tan}^{{n}} \:{x}\:{dx}=\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{tan}^{{n}−\mathrm{1}} \:{x}\:−\int\mathrm{tan}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$ $$\int\mathrm{tan}^{\mathrm{3}} \:{x}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{\mathrm{2}} \:{x}\:−\int\mathrm{tan}\:{x}\:{dx}= \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{\mathrm{2}} \:{x}\:+\mathrm{ln}\:\mathrm{cos}\:{x}\:+{C} \\ $$ $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{tan}^{{n}} \:{x}\:{dx}=+\infty\:\mathrm{for}\:{n}\in\mathbb{N}^{\bigstar} \\ $$ $$\mathrm{what}'\mathrm{s}\:\mathrm{obvious}\:\mathrm{because}\:\mathrm{tan}^{{n}} \:{x}\:\geqslant\mathrm{1}\:\mathrm{for}\:\frac{\pi}{\mathrm{4}}\leqslant{x}<\frac{\pi}{\mathrm{2}} \\ $$ $$\mathrm{and}\:{n}\in\mathbb{N}^{\bigstar} \\ $$

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