Question Number 200778 by sonukgindia last updated on 23/Nov/23
Answered by MM42 last updated on 23/Nov/23
![if n=1⇒∫_0 ^∞ (1/(1+x^2 ))dx =tan^(−1) (x)]_0 ^∞ =(π/2) for n>1 x=tanu⇒∫_0 ^(π/2) ((1+tan^2 u)/((1+tan^2 u)^n ))du=∫_0 ^(π/2) cos^(2n−2) xdx=I_(2n−2) by “expect for expect ” ⇒I_(2n−2) =sinu×cos^(2n−3) u]_0 ^(π/2) +(2n−3)∫_0 ^(π/2) cos^(2n−4) u×(1−cos^2 u)du ⇒I_(2n−2) =((2n−3)/(2n−2)) I_(2n−4) =((2n−3)/(2n−2))×((2n−5)/(2n−4))×...×(1/2)I_2 =((2n−3)/(2n−2))×((2n−5)/(2n−4))×...×(1/2)×(π/2) ✓](Q200796.png)
$${if}\:\:{n}=\mathrm{1}\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\left.={tan}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{2}}\: \\ $$$${for}\:\:{n}>\mathrm{1} \\ $$$${x}={tanu}\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {u}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right)^{{n}} }{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{2}} {xdx}={I}_{\mathrm{2}{n}−\mathrm{2}} \\ $$$${by}\:``{expect}\:\:{for}\:\:{expect}\:'' \\ $$$$\left.\Rightarrow{I}_{\mathrm{2}{n}−\mathrm{2}} ={sinu}×{cos}^{\mathrm{2}{n}−\mathrm{3}} {u}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:+\left(\mathrm{2}{n}−\mathrm{3}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{4}} {u}×\left(\mathrm{1}−{cos}^{\mathrm{2}} {u}\right){du} \\ $$$$\Rightarrow{I}_{\mathrm{2}{n}−\mathrm{2}} =\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}\:{I}_{\mathrm{2}{n}−\mathrm{4}} =\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}×\frac{\mathrm{2}{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{4}}×...×\frac{\mathrm{1}}{\mathrm{2}}{I}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}×\frac{\mathrm{2}{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{4}}×...×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}\:\:\checkmark \\ $$$$ \\ $$