Integration Questions

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

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

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

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

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

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

Answered by ajfour last updated on 06/Aug/19

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

Commented by MJS last updated on 06/Aug/19

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

Answered by MJS last updated on 06/Aug/19

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