Question Number 84578 by msup trace by abdo last updated on 14/Mar/20
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dx}}{\left({cosx}\:+\mathrm{3}{sinx}\right)^{\mathrm{2}} } \\ $$
Commented by jagoll last updated on 14/Mar/20
![∫ _0 ^(π/4) (( dx)/(((√(10)) cos (x−tan^(−1) (3))^2 )) = (1/(10)) ∫ _0 ^(π/4) sec^2 (x−tan^(−1) (3))dx =(1/(10)) tan (x−tan^(−1) (3))] _0^(π/4) = (1/(10)) [tan ((π/4)−tan^(−1) (3))−tan (−tan^(−1) (3))]](https://www.tinkutara.com/question/Q84604.png)
$$\int\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\:}}\frac{\:\mathrm{dx}}{\left(\sqrt{\mathrm{10}}\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\right)\right)^{\mathrm{2}} \right.} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{10}}\:\int\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\:}}\:\mathrm{sec}\:^{\mathrm{2}} \left(\mathrm{x}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\right)\right)\mathrm{dx} \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{10}}\:\mathrm{tan}\:\left(\mathrm{x}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\right)\right)\right]\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{10}}\:\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\right)\right)−\mathrm{tan}\:\left(−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\right)\right)\right] \\ $$$$ \\ $$
Commented by jagoll last updated on 14/Mar/20
![=(1/(10)) [((1−3)/(1+3)) +3 ] = (1/(10)) [(5/2)] = (1/4)](https://www.tinkutara.com/question/Q84605.png)
$$=\frac{\mathrm{1}}{\mathrm{10}}\:\left[\frac{\mathrm{1}−\mathrm{3}}{\mathrm{1}+\mathrm{3}}\:+\mathrm{3}\:\right]\:=\:\frac{\mathrm{1}}{\mathrm{10}}\:\left[\frac{\mathrm{5}}{\mathrm{2}}\right]\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by abdomathmax last updated on 14/Mar/20
![let f(a)=∫_0 ^(π/4) (dx/(a +cosx +3sinx)) ⇒ f^′ (a)=−∫_0 ^(π/4) (dx/((a +cosx +3sinx)^2 )) ⇒ ∫_0 ^(π/4) (dx/((cosx +3sinx)^2 ))=−f^′ (0) let explicit f(a) f(a) =_(tan((x/2))=t) ∫_0 ^((√2)−1) (1/(a+((1−t^2 )/(1+t^2 ))+3((2t)/(1+t^2 ))))((2dt)/(1+t^2 )) = ∫_0 ^((√2)−1) ((2dt)/(a(1+t^2 )+1−t^2 +6t)) =∫_0 ^((√2)−1) (dt/((a−1)t^2 +6t +a+1)) Δ^′ =9−(a−1)(a+1) =9−(a^2 −1)=10−a^2 t_1 =((−3+(√(10−a^2 )))/(a−1)) and t_2 =((−3−(√(10−a^2 )))/(a−1)) (a≠1) f(a) =∫_0 ^((√2)−1) ((2dt)/((a−1)(t−t_1 )(t−t_2 ))) =(2/(a−1)) (1/(t_1 −t_2 ))∫_0 ^((√2)−1) {(1/(t−t_1 ))−(1/(t−t_2 ))}dt =(2/(a−1))×(1/((2(√(10−a^2 )))/(a−1)))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^((√2)−1) =(1/( (√(10−a^2 )))){ln∣(((√2)−1−t_1 )/( (√2)−1−t_2 ))∣−ln∣(t_1 /t_2 )∣} =(1/( (√(10−a^2 )))){ln∣(((√2)−1−((−3+(√(10−a^2 )))/(a−1)))/( (√2)−1+((3+(√(10−a^2 )))/(a−1))))∣ −ln∣((−3+(√(10−a^2 )))/(−3−(√(10−a^2 ))))∣](https://www.tinkutara.com/question/Q84631.png)
$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dx}}{{a}\:+{cosx}\:+\mathrm{3}{sinx}}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{dx}}{\left({a}\:+{cosx}\:+\mathrm{3}{sinx}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{dx}}{\left({cosx}\:+\mathrm{3}{sinx}\right)^{\mathrm{2}} }=−{f}^{'} \left(\mathrm{0}\right)\:{let}\:{explicit}\:{f}\left({a}\right) \\ $$$${f}\left({a}\right)\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{{a}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{6}{t}} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{dt}}{\left({a}−\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{6}{t}\:+{a}+\mathrm{1}} \\ $$$$\Delta^{'} =\mathrm{9}−\left({a}−\mathrm{1}\right)\left({a}+\mathrm{1}\right)\:=\mathrm{9}−\left({a}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{10}−{a}^{\mathrm{2}} \\ $$$${t}_{\mathrm{1}} =\frac{−\mathrm{3}+\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{{a}−\mathrm{1}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{3}−\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{{a}−\mathrm{1}}\:\:\:\left({a}\neq\mathrm{1}\right) \\ $$$${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{2}{dt}}{\left({a}−\mathrm{1}\right)\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{{a}−\mathrm{1}}\:\frac{\mathrm{1}}{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \left\{\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}{dt} \\ $$$$=\frac{\mathrm{2}}{{a}−\mathrm{1}}×\frac{\mathrm{1}}{\frac{\mathrm{2}\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{{a}−\mathrm{1}}}\left[{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}\left\{{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{1}} }{\:\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{2}} }\mid−{ln}\mid\frac{{t}_{\mathrm{1}} }{{t}_{\mathrm{2}} }\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}\left\{{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−\frac{−\mathrm{3}+\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{{a}−\mathrm{1}}}{\:\sqrt{\mathrm{2}}−\mathrm{1}+\frac{\mathrm{3}+\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{{a}−\mathrm{1}}}\mid\right. \\ $$$$−{ln}\mid\frac{−\mathrm{3}+\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{−\mathrm{3}−\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}\mid \\ $$
Commented by abdomathmax last updated on 14/Mar/20
$${f}\left({a}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}\left\{{ln}\mid\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)+\mathrm{3}−\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)+\mathrm{3}+\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}\mid\right. \\ $$$$−{ln}\mid\frac{\mathrm{3}−\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}{\mathrm{3}+\sqrt{\mathrm{10}−{a}^{\mathrm{2}} }}\mid\:{rest}\:{calculus}\:{of}\:{f}^{'} \left({a}\right)\:{and}\:{f}^{'} \left(\mathrm{0}\right) \\ $$$$… \\ $$