Integration Questions

Question Number 87121 by M±th+et£s last updated on 03/Apr/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\$$

Answered by redmiiuser last updated on 03/Apr/20

$$\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{−\mathrm{1}} \\$$ $$=\underset{{n}=\mathrm{0}\:\:} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}} \\$$ $$\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{−\mathrm{1}} \\$$ $$=\left(\mathrm{1}−{x}^{\mathrm{4}} \right).\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}} \right) \\$$ $$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}} −\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}+\mathrm{4}} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\$$ $$=\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{4}{n}+\mathrm{1}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\right]}}−\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}+\mathrm{5}} }{\mathrm{4}{n}+\mathrm{5}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\right]}} \\$$ $$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{4}{n}+\mathrm{1}}\:\:−\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}{n}+\mathrm{5}} }{\mathrm{4}{n}+\mathrm{5}} \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${pls}\:{pls}\:{pls}\:{check} \\$$

Commented byM±th+et£s last updated on 03/Apr/20

$${its}\:{right}\:{sir}\:{thank}\:{you} \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${welcome}\:{mister} \\$$

Commented byMJS last updated on 03/Apr/20

$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{your}\:\mathrm{formula}? \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${from}\:{the}\:{closed}\:{forms} \\$$ $${its}\:{much}\:{efficent}\:{to} \\$$ $${insert}\:\:{the}\:{borders} \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${Anyway}\:{Thanks}\:{Sir}! \\$$

Commented bymr W last updated on 03/Apr/20

$${as}\:{much}\:{as}\:{one}\:{can}\:{not}\:{get}\:{the}\:{value} \\$$ $${of}\:{the}\:{infinite}\:{series},\:{i}\:{think}\:{such}\:{a} \\$$ $${solution}\:{is}\:{not}\:{so}\:{useful}. \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${The}\:{process}\:{is}\:{not}\: \\$$ $${a}\:{closed}\:{form}\:{but}\:{yet} \\$$ $${it}\:{is}\:{efficent}. \\$$ $${Anyway}\:{Thanks}\:{for} \\$$ $${your}\:{comment}. \\$$

Answered by MJS last updated on 03/Apr/20

$$\int\frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\int\left(\frac{\mathrm{2}}{{x}^{\mathrm{4}} +\mathrm{1}}−{x}\right){dx}= \\$$ $$=−\int{dx}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}= \\$$ $$=−{x}− \\$$ $$\:\:\:\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+ \\$$ $$\:\:\:\:\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+{C}= \\$$ $$=−{x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\mathrm{2}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right)\right)\:+{C}= \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${sir}\:{its}\:{a}\:{definite} \\$$ $${integration}. \\$$

Commented byMJS last updated on 03/Apr/20

$$\mathrm{you}\:\mathrm{can}\:\mathrm{insert}\:\mathrm{the}\:\mathrm{borders} \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${sir}\:{can}\:{you}\:{make}\:{the} \\$$ $${process}\:{more}\:{smooth}. \\$$

Commented byM±th+et£s last updated on 03/Apr/20

$${thank}\:{you}\:{its}\:{easy}\:{to}\:{insert}\:{the}\:{borders}\:{now} \\$$

Commented byredmiiuser last updated on 03/Apr/20

$${God}\:{bless}\:{you}\:\:{Sir}. \\$$ $${Have}\:{a}\:{nice}\:{day}. \\$$

Answered by TANMAY PANACEA. last updated on 03/Apr/20

$$\left(−\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{4}} +\mathrm{1}−\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{4}} }−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\$$ $$\mid\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}}\right)−{x}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$ $$\left\{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}}{\pi}}{\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\frac{\pi}{\mathrm{2}}+\frac{\mathrm{2}}{\pi}−\sqrt{\mathrm{2}}}{\frac{\pi}{\mathrm{2}}+\frac{\mathrm{2}}{\pi}+\sqrt{\mathrm{2}}}\right)−\frac{\pi}{\mathrm{2}}\right\}− \\$$ $$\\$$ $$\left\{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(−\infty\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{0}+\mathrm{1}−\mathrm{0}}{\mathrm{0}+\mathrm{1}−\mathrm{0}}\right)−\mathrm{0}\right\} \\$$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\pi^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{2}}\:\pi}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\pi^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\:\pi}{\pi^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\pi}\right)−\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\frac{−\pi}{\mathrm{2}}\right) \\$$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\pi^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{2}\pi}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\pi^{\mathrm{2}} +\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\:\pi}{\pi^{\mathrm{2}} +\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\:\pi}\right)−\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right) \\$$ $$\bigstar{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}}\right)\rightarrow{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{2}}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}}\:{x}}\right)\:\bigstar \\$$ $$\\$$

Commented byTANMAY PANACEA. last updated on 03/Apr/20

$${most}\:{welcome}\:{sir}... \\$$

Commented byM±th+et£s last updated on 03/Apr/20

$${thank}\:{you}\:{sir} \\$$

Commented bypeter frank last updated on 03/Apr/20

$${thank}\:{you}\:{both} \\$$