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 149023 by ArielVyny last updated on 02/Aug/21

∫_0 ^(π/4) ((tsint)/(1+cos^2 t))dt=∫_0 ^(π/4) ((tcost)/(1+sin^2 t))dt  true or false ??

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{tsint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}=\overset{\frac{\pi}{\mathrm{4}}} {\int}_{\mathrm{0}} \frac{{tcost}}{\mathrm{1}+{sin}^{\mathrm{2}} {t}}{dt} \\ $$$${true}\:{or}\:{false}\:?? \\ $$

Answered by mindispower last updated on 02/Aug/21

((sin(t))/(1+cos^2 (t)))<((cos(t))/(1+sin^2 (t))),∀t∈[0,(π/4)[  proff  cos(t)>sin(t)≥0⇒  (1/(1+cos^2 (t)))<(1/(1+sin^2 (t)))⇒((sin(t))/(1+cos^2 (t)))<((cos(t))/(1+sin^2 (t)))  ⇒0≤t((sin(t))/(1+cos^2 (t)))<((tcos(t))/(1+sin^2 (t)))  ∫_0 ^(π/4) ((tsin(t))/(1+cos^2 (t)))dt<∫_0 ^(π/4) ((tcos(t))/(1+sin^2 (t)))dt

$$\frac{{sin}\left({t}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({t}\right)}<\frac{{cos}\left({t}\right)}{\mathrm{1}+{sin}^{\mathrm{2}} \left({t}\right)},\forall{t}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\left[\right.\right. \\ $$$${proff}\:\:{cos}\left({t}\right)>{sin}\left({t}\right)\geqslant\mathrm{0}\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({t}\right)}<\frac{\mathrm{1}}{\mathrm{1}+{sin}^{\mathrm{2}} \left({t}\right)}\Rightarrow\frac{{sin}\left({t}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({t}\right)}<\frac{{cos}\left({t}\right)}{\mathrm{1}+{sin}^{\mathrm{2}} \left({t}\right)} \\ $$$$\Rightarrow\mathrm{0}\leqslant{t}\frac{{sin}\left({t}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({t}\right)}<\frac{{tcos}\left({t}\right)}{\mathrm{1}+{sin}^{\mathrm{2}} \left({t}\right)} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{tsin}\left({t}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({t}\right)}{dt}<\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{tcos}\left({t}\right)}{\mathrm{1}+{sin}^{\mathrm{2}} \left({t}\right)}{dt} \\ $$$$ \\ $$

Commented by mindispower last updated on 03/Aug/21

yes withe special function

$${yes}\:{withe}\:{special}\:{function} \\ $$

Commented by ArielVyny last updated on 02/Aug/21

sir you can find the value of this integral?

$${sir}\:{you}\:{can}\:{find}\:{the}\:{value}\:{of}\:{this}\:{integral}? \\ $$

Commented by ArielVyny last updated on 06/Aug/21

you can solve it please

$${you}\:{can}\:{solve}\:{it}\:{please} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com