Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 336 by Vishal Bhardwaj last updated on 25/Jan/15

∫ (1/(sin^4 x+cos^4 x+sin^2 x cos^2 x)) dx

$$\int\:\frac{\mathrm{1}}{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{2}} {x}\:{cos}^{\mathrm{2}} {x}}\:\:{dx} \\ $$

Commented by 123456 last updated on 22/Dec/14

f(x)=(1/(sin^4 x+cos^4 x+sin^2 x cos^2 x)) =(1/(sin^4 x+cos^4 x+2sin^2 x cos^2 x−sin^2 x cos^2 x)) =(1/((sin^2 x+cos^2 x)^2 −sin^2 x cos^2 x)) =(1/(1−sin^2 x cos^2 x))

$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{2sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}} \\ $$

Commented by 123456 last updated on 23/Dec/14

((tan^(−1) [((√3)/2)tan(2x)])/(√3))

$$\frac{\mathrm{tan}^{−\mathrm{1}} \left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}\left(\mathrm{2}{x}\right)\right]}{\sqrt{\mathrm{3}}} \\ $$

Answered by prakash jain last updated on 27/Dec/14

(1/(1−sin^2 xcos^2 x))=(4/(4−sin^2 2x)) tan2x=t sec^2 2x∙2∙dx=dt dx=(dt/(2(1+t^2 ))) sin^2 2x=(t^2 /(1+t^2 )) Given integral I I=∫(((4dt)/(2(1+t^2 )))/(4−(t^2 /((1+t^2 ))))) =∫ ((4dt)/(2(4+4t^2 −t^2 ))) =2∫ (dt/(3t^2 +4)) =2∫ (dt/(t^2 +(4/3)))=(2/3)∫ (dt/(t^2 +((2/(√3)))^2 )) =(2/3)∙((√3)/2)tan^(−1) (((√3)t)/2) =(1/(√3))tan^(−1) (((√3)tan 2x)/2)

$$\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\mathrm{cos}^{\mathrm{2}} {x}}=\frac{\mathrm{4}}{\mathrm{4}−\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}} \\ $$$$\mathrm{tan2}{x}={t} \\ $$$$\mathrm{sec}^{\mathrm{2}} \mathrm{2}{x}\centerdot\mathrm{2}\centerdot{dx}={dt} \\ $$$${dx}=\frac{{dt}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}=\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{Given}\:\mathrm{integral}\:{I} \\ $$$${I}=\int\frac{\frac{\mathrm{4}{dt}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}}{\mathrm{4}−\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}} \\ $$$$=\int\:\frac{\mathrm{4}{dt}}{\mathrm{2}\left(\mathrm{4}+\mathrm{4}{t}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}\int\:\frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}} \\ $$$$=\mathrm{2}\int\:\frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}}}=\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{{dt}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}{t}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com