Question Number 219341 by alcohol last updated on 23/Apr/25
$$\int\frac{{dx}}{\mathrm{1}\:+\:{sin}^{\mathrm{3}} {x}\:+\:{cos}^{\mathrm{3}} {x}} \\ $$
Commented by Ghisom last updated on 23/Apr/25
$$\mathrm{simply}\:\mathrm{use}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{decompose} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$${I}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\:\left({I}_{{k}} \right)\:+{C}\: \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{9}}\int\frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}_{\mathrm{3}} =−\frac{\mathrm{4}}{\mathrm{9}}\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$${I}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\int{dt} \\ $$$$\mathrm{these}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$
Answered by Ghisom last updated on 23/Apr/25
![if you don′t like (x/2) in the answer: ∫(dx/(1+cos^3 x +sin^3 x))= =∫(((1+tan^2 x)^(3/2) )/(1+tan^3 x +(1+tan^2 x)^(3/2) ))dx= [ _(x=arctan ((t^2 −1)/(2t)))^(t=tan x +(√(1+tan^2 x))) → dx=((2dt)/(t^2 +1))] =∫(((t^2 +1)^2 )/(t^2 (t+1)^2 (t^2 −2t+3)))dt= =I=Σ_(k=1) ^5 I_k I_1 =(4/9)∫(dt/(t^2 −2t+3))=((2(√2))/9)arctan (((√2)(t−1))/2) I_2 =(2/3)∫(dt/((t+1)^2 ))=−(2/(3(t+1))) I_3 =(1/3)∫(dt/t^2 )=−(1/(3t)) I_4 =(4/9)∫(dt/(t+1))=(4/9)ln (t+1) I_5 =−(4/9)∫(dt/t)=−(4/9)ln t I=−((3t+1)/(3t(t+1)))+(4/9)ln ((t+1)/t) +((2(√2))/9)arctan (((√2)(t−1))/2) now insert t=((1+sin x)/(cos x))](https://www.tinkutara.com/question/Q219376.png)
$$\mathrm{if}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{like}\:\frac{{x}}{\mathrm{2}}\:\mathrm{in}\:\mathrm{the}\:\mathrm{answer}: \\ $$$$\int\frac{{dx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{3}} \:{x}\:+\mathrm{sin}^{\mathrm{3}} \:{x}}= \\ $$$$=\int\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{1}+\mathrm{tan}^{\mathrm{3}} \:{x}\:+\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)^{\mathrm{3}/\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[\:_{{x}=\mathrm{arctan}\:\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}}} ^{{t}=\mathrm{tan}\:{x}\:+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}} \:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{3}\right)}{dt}= \\ $$$$={I}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}{I}_{{k}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{9}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\left({t}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{2}}{\mathrm{3}\left({t}+\mathrm{1}\right)} \\ $$$${I}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}{t}} \\ $$$${I}_{\mathrm{4}} =\frac{\mathrm{4}}{\mathrm{9}}\int\frac{{dt}}{{t}+\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{9}}\mathrm{ln}\:\left({t}+\mathrm{1}\right) \\ $$$${I}_{\mathrm{5}} =−\frac{\mathrm{4}}{\mathrm{9}}\int\frac{{dt}}{{t}}=−\frac{\mathrm{4}}{\mathrm{9}}\mathrm{ln}\:{t} \\ $$$${I}=−\frac{\mathrm{3}{t}+\mathrm{1}}{\mathrm{3}{t}\left({t}+\mathrm{1}\right)}+\frac{\mathrm{4}}{\mathrm{9}}\mathrm{ln}\:\frac{{t}+\mathrm{1}}{{t}}\:+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\left({t}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{insert}\:{t}=\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$