Question Number 225939 by Spillover last updated on 16/Nov/25
Answered by Ghisom_ last updated on 16/Nov/25
![∫_0 ^(π/4) sin^2 x (√(tan x)) dx= [t=(√(cot x)) → dx=−2sin^2 x (√(cot x))] =−2∫_∞ ^1 (dt/((t^4 +1)^2 ))=2∫_1 ^∞ (dt/((t^4 +1)^2 ))= [Ostrogradski′s Method] =[(t/(2(t^4 +1)))]_1 ^∞ +(3/2)∫_1 ^∞ (dt/(t^4 +1))= [decompose t^4 +1=(t^2 −(√2)t+1)(t^2 +(√2)t+1] =−(1/4)+((3(√2))/8)ln ((√2)−1) +((3(√2))/(16))π≈.115621620800](https://www.tinkutara.com/question/Q225948.png)
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{cot}\:{x}}\:\rightarrow\:{dx}=−\mathrm{2sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cot}\:{x}}\right] \\ $$$$=−\mathrm{2}\underset{\infty} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dt}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\left[\frac{{t}}{\mathrm{2}\left({t}^{\mathrm{4}} +\mathrm{1}\right)}\right]_{\mathrm{1}} ^{\infty} +\frac{\mathrm{3}}{\mathrm{2}}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}}= \\ $$$$\:\:\:\:\:\left[\mathrm{decompose}\:{t}^{\mathrm{4}} +\mathrm{1}=\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right]\right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:+\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{16}}\pi\approx.\mathrm{115621620800} \\ $$
Answered by Spillover last updated on 17/Nov/25
