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Question Number 36941 by maxmathsup by imad last updated on 07/Jun/18
calculate ∫_0 ^((3π)/4) (dt/((1+sin^2 t)^2 ))
$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} } \\ $$
Commented by math khazana by abdo last updated on 08/Jun/18
we have cos^2 t = (1/(1+tan^2 t)) ⇒sin^2 t =1−(1/(1+tan^2 t)) = ((tan^2 t)/(1+tan^2 t)) ⇒ I = ∫_0 ^((3π)/4) (dt/((1+((tan^2 t)/(1+tan^2 t)))^2 )) =_(tant =−x) ∫_0 ^1 (1/((1+(x^2 /(1+x^2 )))^2 )) .((−dx)/(1+x^2 )) =− ∫_0 ^1 (((1+x^2 )^2 )/((1+x^2 )(1+2x^2 )^2 ))dx =−∫_0 ^1 ((1+x^2 )/((1+2x^2 )^2 ))dx changement x(√2) = sh(u) give I = −∫_0 ^(argsh((√2))) ((1+(1/2)sh^2 u)/((ch^4 u))) (1/( (√2))) ch(u)du =− (1/(2(√2))) ∫_0 ^(ln((√2) +(√3))) ((2 +sh^2 u)/(ch^3 u)) du =−(1/(2(√2))) ∫_0 ^(ln((√2) +(√3))) ((2 + ((ch(2u)−1)/2))/(((1+ch(2u))/2) ch(u))) du = (1/(2(√2))) ∫_0 ^(ln((√2) +(√3))) ((3+ch(2u))/(chu +ch(u)ch(2u))) du... be continued...
$${we}\:{have}\:{cos}^{\mathrm{2}} {t}\:=\:\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\:\Rightarrow{sin}^{\mathrm{2}} {t}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}} \\ $$$$=\:\frac{{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\:\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+\frac{{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\right)^{\mathrm{2}} } \\ $$$$=_{{tant}\:=−{x}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:.\frac{−{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:{changement}\:{x}\sqrt{\mathrm{2}}\:=\:{sh}\left({u}\right) \\ $$$${give}\:\:{I}\:=\:−\int_{\mathrm{0}} ^{{argsh}\left(\sqrt{\mathrm{2}}\right)} \:\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{sh}^{\mathrm{2}} {u}}{\left({ch}^{\mathrm{4}} {u}\right)}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{ch}\left({u}\right){du} \\ $$$$=−\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\int_{\mathrm{0}} ^{{ln}\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\right)} \:\:\:\:\:\frac{\mathrm{2}\:+{sh}^{\mathrm{2}} {u}}{{ch}^{\mathrm{3}} {u}}\:{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{{ln}\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\right)} \:\:\frac{\mathrm{2}\:+\:\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\:{ch}\left({u}\right)}\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{{ln}\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\right)} \:\:\frac{\mathrm{3}+{ch}\left(\mathrm{2}{u}\right)}{{chu}\:+{ch}\left({u}\right){ch}\left(\mathrm{2}{u}\right)}\:{du}… \\ $$$${be}\:{continued}… \\ $$$$ \\ $$

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