Question Number 202224 by sonukgindia last updated on 23/Dec/23
Answered by witcher3 last updated on 23/Dec/23
![x→(1/(a+bcos^2 (x)))=f(x),is p peridic ⇒∫_(kπ) ^((k+1)π) f(x)dx=∫_0 ^π f(x)dx;by kπ+y=x ⇒∫_0 ^(nπ) f(x)dx=Σ_(k=0) ^(n−1) ∫_((kπ)) ^((k+1)π) f(x)dx=Σ_(k=0) ^(n−1) ∫_0 ^π f(x)dx =n∫_0 ^π (dx/(a+bcos^2 (x)))=n(∫_0 ^(π/2) (dx/(a+bcos^2 (x)))+∫_(π/2) ^π (dx/(a+bcos^2 (x))) _(π−x=h) ) =2n∫_0 ^(π/2) (dx/(a+bcos^2 (x)))=((2n)/b)∫_0 ^(π/2) (1/(cos^2 (x)))((1/(1+(a/b)(1+tan^2 (x))))) =((2n)/b)∫_0 ^(π/2) ((d(tan(x)))/((((b+a)/b))(1+(tan(x).(√(a/(b+a))))^2 )) =2n(√(1/(a(b+a))))∫_0 ^(π/2) .((d((√(a/(b+a))).tan(x)))/(1+((√(a/(b+a))).tan(x))^2 ));(a/(b+a))>0,b+a#0 =2n(√((1/(a(b+a))).))tan^(−1) ((√(a/(b+a))).tan (x)]_0 ^(π/2) =((πn)/( (√(a(a+b)))))](https://www.tinkutara.com/question/Q202234.png)
$$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}=\mathrm{f}\left(\mathrm{x}\right),\mathrm{is}\:\mathrm{p}\:\mathrm{peridic} \\ $$$$\Rightarrow\int_{\mathrm{k}\pi} ^{\left(\mathrm{k}+\mathrm{1}\right)\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx};\mathrm{by}\:\mathrm{k}\pi+\mathrm{y}=\mathrm{x} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{n}\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\int_{\left(\mathrm{k}\pi\right)} ^{\left(\mathrm{k}+\mathrm{1}\right)\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\mathrm{n}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{dx}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}=\mathrm{n}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \frac{\mathrm{dx}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}\:_{\pi−\mathrm{x}=\mathrm{h}} \right) \\ $$$$=\mathrm{2n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}=\frac{\mathrm{2n}}{\mathrm{b}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{a}}{\mathrm{b}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)\right)}\right) \\ $$$$=\frac{\mathrm{2n}}{\mathrm{b}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{d}\left(\mathrm{tan}\left(\mathrm{x}\right)\right)}{\left(\frac{\mathrm{b}+\mathrm{a}}{\mathrm{b}}\right)\left(\mathrm{1}+\left(\mathrm{tan}\left(\mathrm{x}\right).\sqrt{\frac{\mathrm{a}}{\mathrm{b}+\mathrm{a}}}\right)^{\mathrm{2}} \right.} \\ $$$$=\mathrm{2n}\sqrt{\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{b}+\mathrm{a}\right)}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} .\frac{\mathrm{d}\left(\sqrt{\frac{\mathrm{a}}{\mathrm{b}+\mathrm{a}}}.\mathrm{tan}\left(\mathrm{x}\right)\right)}{\mathrm{1}+\left(\sqrt{\frac{\mathrm{a}}{\mathrm{b}+\mathrm{a}}}.\mathrm{tan}\left(\mathrm{x}\right)\right)^{\mathrm{2}} };\frac{\mathrm{a}}{\mathrm{b}+\mathrm{a}}>\mathrm{0},\mathrm{b}+\mathrm{a}#\mathrm{0} \\ $$$$=\mathrm{2n}\sqrt{\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{b}+\mathrm{a}\right)}.}\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{a}}{\mathrm{b}+\mathrm{a}}}.\mathrm{tan}\:\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi\mathrm{n}}{\:\sqrt{\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)}} \\ $$
Answered by Calculusboy last updated on 23/Dec/23
![Solution: multiply both numerator and denominator by (1/(cos^2 x)) I=∫_0 ^(n𝛑) ((1/(cos^2 x))/((a/(cos^2 x))+((bsin^2 x)/(cos^2 x))))dx NB: [ ((sinx)/(cosx))=tanx and (1/(cosx))=secx] I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(asec^2 x+btan^2 x))dx NB: sec^2 x=1+tan^2 x I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(a(1+tan^2 x)+btan^2 x))dx I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(a+atan^2 x+btan^2 x))dx ⇔ I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(a+tan^2 x(a+b)))dx let u=tanx du=sec^2 xdx [when x=(𝛑/2) u=∞ and when x=0 u=0] I=2n∫_0 ^∞ ((sec^2 x)/(a+u^2 (a+b)))∙(du/(sec^2 x)) I=2n∫_0 ^∞ (du/(u^2 (a+b)+a)) ⇔ I=((2n)/((a+b)))∫_0 ^∞ (du/(u^2 +((√(a/((a+b)))))^2 )) I=((2n)/((a+b)))×(1/( (√(a/((a+b)))))) ∣tan^(−1) [(u/( (√(a/((a+b))))))]∣_0 ^∞ I={((2n)/((a+b)))×((√(a+b))/( (√a)))}∣tan^(−1) [((u(√((a+b))))/( (√a)))]∣_0 ^∞ I=((2n(√((a+b))))/( (√a)(a+b)))×(𝛑/2)=((n𝛑(√((a+b))))/( (√a)(a+b)))](https://www.tinkutara.com/question/Q202238.png)
$$\boldsymbol{{Solution}}:\:\:\boldsymbol{{multiply}}\:\boldsymbol{{both}}\:\boldsymbol{{numerator}}\:\boldsymbol{{and}}\:\boldsymbol{{denominator}}\:\boldsymbol{{by}}\:\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\boldsymbol{{n}\pi}} \:\frac{\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}}{\frac{\boldsymbol{{a}}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}+\frac{\boldsymbol{{bsin}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}}\boldsymbol{{dx}}\:\:\:\boldsymbol{{NB}}:\:\left[\:\frac{\boldsymbol{{sinx}}}{\boldsymbol{{cosx}}}=\boldsymbol{{tanx}}\:\:\boldsymbol{{and}}\:\frac{\mathrm{1}}{\boldsymbol{{cosx}}}=\boldsymbol{{secx}}\right] \\ $$$$\boldsymbol{{I}}=\mathrm{2}\boldsymbol{{n}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{asec}}^{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{{btan}}^{\mathrm{2}} \boldsymbol{{x}}}\boldsymbol{{dx}}\:\:\:\boldsymbol{{NB}}:\:\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}=\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}} \\ $$$$\boldsymbol{{I}}=\mathrm{2}\boldsymbol{{n}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{a}}\left(\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}\right)+\boldsymbol{{btan}}^{\mathrm{2}} \boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\mathrm{2}\boldsymbol{{n}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{a}}+\boldsymbol{{atan}}^{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{{btan}}^{\mathrm{2}} \boldsymbol{{x}}}\boldsymbol{{dx}}\:\Leftrightarrow\:\boldsymbol{{I}}=\mathrm{2}\boldsymbol{{n}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{a}}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{tanx}}\:\:\boldsymbol{{du}}=\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{xdx}}\:\:\:\left[\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\boldsymbol{{u}}=\infty\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{{u}}=\mathrm{0}\right] \\ $$$$\boldsymbol{{I}}=\mathrm{2}\boldsymbol{{n}}\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{a}}+\boldsymbol{{u}}^{\mathrm{2}} \left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}\centerdot\frac{\boldsymbol{{du}}}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}} \\ $$$$\boldsymbol{{I}}=\mathrm{2}\boldsymbol{{n}}\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{du}}}{\boldsymbol{{u}}^{\mathrm{2}} \left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)+\boldsymbol{{a}}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\frac{\mathrm{2}\boldsymbol{{n}}}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{du}}}{\boldsymbol{{u}}^{\mathrm{2}} +\left(\sqrt{\frac{\boldsymbol{{a}}}{\left(\boldsymbol{{a}}+\boldsymbol{\mathrm{b}}\right)}}\right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{2}\boldsymbol{{n}}}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}×\frac{\mathrm{1}}{\:\sqrt{\frac{\boldsymbol{{a}}}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}}}\:\mid\boldsymbol{{tan}}^{−\mathrm{1}} \left[\frac{\boldsymbol{{u}}}{\:\sqrt{\frac{\boldsymbol{{a}}}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}}}\right]\mid_{\mathrm{0}} ^{\infty} \\ $$$$\boldsymbol{{I}}=\left\{\frac{\mathrm{2}\boldsymbol{{n}}}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}×\frac{\sqrt{\boldsymbol{{a}}+\boldsymbol{{b}}}}{\:\sqrt{\boldsymbol{{a}}}}\right\}\mid\boldsymbol{{tan}}^{−\mathrm{1}} \left[\frac{\boldsymbol{{u}}\sqrt{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}}{\:\sqrt{\boldsymbol{{a}}}}\right]\mid_{\mathrm{0}} ^{\infty} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{2}\boldsymbol{{n}}\sqrt{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}}{\:\sqrt{\boldsymbol{{a}}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}×\frac{\boldsymbol{\pi}}{\mathrm{2}}=\frac{\boldsymbol{{n}\pi}\sqrt{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}}{\:\sqrt{\boldsymbol{{a}}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)} \\ $$$$ \\ $$