Question Number 83206 by mathmax by abdo last updated on 28/Feb/20
$$\left.\mathrm{1}\right)\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{dx}}{\mathrm{2}+{a}\:{sinx}}\:\:\:\:\left({areal}\right) \\ $$$$\left.\mathrm{2}\right)\:{c}\:{explicite}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sinx}}{\left(\mathrm{2}+{asinx}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 29/Feb/20
![1) let f(a)=∫_0 ^(π/4) (dx/(2+asinx)) changement tan((x/2))=t give f(a) =∫_0 ^((√2)−1) ((2dt)/((1+t^2 )(2+a((2t)/(1+t^2 ))))) =2∫_0 ^((√2)−1) (dt/(2+2t^2 +2at)) =∫_0 ^((√2)−1) (dt/(t^2 +at +1)) t^2 +at+1=0→Δ=a^2 −4 if ∣a∣>2 ⇒ Δ>0 ⇒t_1 =((−a+(√(a^2 −4)))/2) and t_2 =((−a−(√(a^2 −4)))/2) ⇒f(a) =∫_0 ^((√2)−1) (dt/((t−t_1 )(t−t_2 ))) =(1/(t−t_2 ))∫_0 ^((√2)−1) ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/( (√(a^2 −4)))) [ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^((√2)−1) =(1/( (√(a^2 −4)))){ln∣(((√2)−1−((−a+(√(a^2 −4)))/2))/( (√2)−1−((−a−(√(a^2 −4)))/2)))∣ −ln∣((−a+(√(a^2 −4)))/(−a−(√(a^2 −4))))∣}=(1/( (√(a^2 −4)))){ln∣((2(√2)−2+a−(√(a^2 −4)))/(2(√2)−2+a+(√(a^2 −4))))∣ −ln∣((a−(√(a^2 −4)))/(a+(√(a^2 −4))))∣ if ∣a∣<2 ⇒Δ<0 ⇒f(a)=∫_0 ^((√2)−1) (dt/(t^2 +((2at)/2) +(a^2 /4)+1−(a^2 /4))) =∫_0 ^((√2)−1) (dt/((t+(a/2))^2 +((4−a^2 )/4))) =_(t+(a/2)=((√(4−a^2 ))/2)u→u=((2t+a)/( (√(4−a^2 ))))) (4/(4−a^2 )) ∫_(a/( (√(4−a^2 )))) ^((2(√2)−2+a)/( (√(4−a^2 )))) (1/(1+u^2 ))×((√(4−a^2 ))/2)du =(2/( (√(4−a^2 ))))[arctanu]_(a/( (√(4−a^2 )))) ^((2(√2)−2+a)/( (√(4−a^2 )))) =(2/( (√(4−a^2 )))){ arctan(((2(√2)−2+a)/( (√(4−a^2 ))))) −arctan((a/( (√(4−a^2 )))))}](https://www.tinkutara.com/question/Q83299.png)
$$\left.\mathrm{1}\right)\:{let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{dx}}{\mathrm{2}+{asinx}}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{2}+{a}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)}\:=\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{at}} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+{at}\:+\mathrm{1}} \\ $$$${t}^{\mathrm{2}} \:+{at}+\mathrm{1}=\mathrm{0}\rightarrow\Delta={a}^{\mathrm{2}} −\mathrm{4} \\ $$$${if}\:\mid{a}\mid>\mathrm{2}\:\Rightarrow\:\Delta>\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\:{and}\:{t}_{\mathrm{2}} =\frac{−{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{{dt}}{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\:\left[{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\left\{{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−\frac{−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}}{\:\sqrt{\mathrm{2}}−\mathrm{1}−\frac{−{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}}\mid\right. \\ $$$$\left.−{ln}\mid\frac{−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{−{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\mid\right\}=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\left\{{ln}\mid\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}+{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}+{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\mid\right. \\ $$$$−{ln}\mid\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\mid \\ $$$${if}\:\mid{a}\mid<\mathrm{2}\:\Rightarrow\Delta<\mathrm{0}\:\Rightarrow{f}\left({a}\right)=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{\mathrm{2}{at}}{\mathrm{2}}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{{dt}}{\left({t}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}−{a}^{\mathrm{2}} }{\mathrm{4}}}\:=_{{t}+\frac{{a}}{\mathrm{2}}=\frac{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}{\mathrm{2}}{u}\rightarrow{u}=\frac{\mathrm{2}{t}+{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}} \frac{\mathrm{4}}{\mathrm{4}−{a}^{\mathrm{2}} }\:\:\:\int_{\frac{{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}} ^{\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}+{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }×\frac{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}{\mathrm{2}}{du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\left[{arctanu}\right]_{\frac{{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}} ^{\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}+{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}} \:\:\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\left\{\:{arctan}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}+{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right)\right. \\ $$$$\left.−{arctan}\left(\frac{{a}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right)\right\} \\ $$
Commented by mathmax by abdo last updated on 29/Feb/20
$$\left.\mathrm{2}\right){we}\:{have}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{2}+{asinx}}\:\Rightarrow{f}^{'} \left({a}\right)=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sinx}\:{dx}}{\left(\mathrm{2}+{asinx}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sinx}\:{dx}}{\left(\mathrm{2}+{asinx}\right)^{\mathrm{2}} }\:=−{f}^{'} \left({a}\right)\:\:{rest}\:{the}\:{calculus}\:{of}\:{f}^{'} \left({a}\right)…{be}\:{continued}… \\ $$