Question Number 211908 by Spillover last updated on 23/Sep/24
Answered by BHOOPENDRA last updated on 24/Sep/24
![I=∫_0 ^(√3) (Σ_(k=0) ^∞ sin (((kπ)/2))x^k )dx The pattern for sin (kπ/2) for k=0 sin (((kπ)/2))=sin (0)=0 For k=1 sin (1×(π/2))=sin (π/2)=1 For k=2 sin (2×(π/2))=sin (π)=0 For k=3 sin (((3π)/2))=−1, For k=4 sin(((4π)/2))=0 For k=5 sin (((5π)/2))=1 S(x)=(x−x^3 +x^5 −x^7 +x^9 +..........) S(x)=Σ_(k=0) ^∞ sin(((kπ)/2))x^(k ) =(x/((1+x^2 ))) I=∫_0 ^(√3) S(x)dx =∫_0 ^(√3) (x/(1+x^2 )) dx [ln(1+x^2 )/2]_0 ^(√3) =((ln(4))/2)= ln2](https://www.tinkutara.com/question/Q211912.png)
$${I}=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{sin}\:\left(\frac{{k}\pi}{\mathrm{2}}\right){x}^{{k}} \right){dx} \\ $$$${The}\:{pattern}\:{for}\:\mathrm{sin}\:\left({k}\pi/\mathrm{2}\right) \\ $$$${for}\:{k}=\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\frac{{k}\pi}{\mathrm{2}}\right)=\mathrm{sin}\:\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${For}\:{k}=\mathrm{1} \\ $$$$\mathrm{sin}\:\left(\mathrm{1}×\frac{\pi}{\mathrm{2}}\right)=\mathrm{sin}\:\left(\pi/\mathrm{2}\right)=\mathrm{1} \\ $$$${For}\:{k}=\mathrm{2} \\ $$$$\mathrm{sin}\:\left(\mathrm{2}×\frac{\pi}{\mathrm{2}}\right)=\mathrm{sin}\:\left(\pi\right)=\mathrm{0} \\ $$$${For}\:{k}=\mathrm{3} \\ $$$$\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)=−\mathrm{1},\:{For}\:{k}=\mathrm{4}\:\:\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${For}\:{k}=\mathrm{5}\: \\ $$$$\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{2}}\right)=\mathrm{1} \\ $$$${S}\left({x}\right)=\left({x}−{x}^{\mathrm{3}} +{x}^{\mathrm{5}} −{x}^{\mathrm{7}} +{x}^{\mathrm{9}} +……….\right) \\ $$$$\:\:\:\:\:\:\:{S}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right){x}^{{k}\:} =\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${I}=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} {S}\left({x}\right){dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\left[{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)/\mathrm{2}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \\ $$$$=\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{2}}=\:{ln}\mathrm{2} \\ $$
Commented by Spillover last updated on 24/Sep/24
$${good}.{thanks} \\ $$