Limits and Integrals!

Calculus Level 4

$\displaystyle \lim_{n\rightarrow\infty}\left(\dfrac{2n}{n^{2}+1}+\dfrac{2n}{n^{2}+4}+\dfrac{2n}{n^{2}+9}+\cdots+\dfrac{2n}{n^{2}+n^{2}}\right)$

Let $A$ denote the value of the limit above. And if the value of, $\displaystyle I=\int_{0}^{A}\dfrac{x}{\sin(x)+\cos(x)}dx$

is of the form, $\displaystyle \dfrac{\pi}{a\sqrt{b}}\left(\ln(\sqrt{c}+d)-\ln(\sqrt{e}-f)\right)$

Find the value of $a+b+c+d+e+f$ .

Details and Assumptions :

1) $a,b,c,d,e,f$ are integers . They need not to be distinct

2) $b,c,e$ are not having any factor which is a perfect square.

1 solution

Tanishq Varshney
May 3, 2015

using reimann sum form

$A=2 \displaystyle \int^{1}_{0} \frac{dx}{1+x^{2}}$

$A=\frac{\pi}{2}$

$I=\displaystyle \int^{\frac{\pi}{2}}_{0} \frac{x}{sinx+cosx}dx$

using property of definite integral

$2I=\frac{\pi}{2} \displaystyle \int^{\frac{\pi}{2}}_{0} \frac{1}{sinx+cosx}dx$

$\large{sinx=\frac{tan(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})}}$

$\large{cosx=\frac{1-tan^{2}(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})}}$

$2I=\frac{\pi}{2} \displaystyle \int^{\frac{\pi}{2}}_{0} \frac{sec^{2}(\frac{x}{2})}{2tan(\frac{x}{2})+1-tan^{2}(\frac{x}{2})}dx$

put $tan(\frac{x}{2})=t$

$2I=\frac{\pi}{2} \displaystyle \int^{1}_{0} \frac{2}{2t+1-t^{2}}dt$

$2I=\pi \displaystyle \int^{1}_{0} \frac{1}{2-(t-1)^{2}}dt$

$I=\frac{\pi}{4\sqrt{2}}(ln(\sqrt{2}+1)-ln(\sqrt{2}-1))$

$a=4,~b=2,~c=2,~d=1,~e=2,~f=1$

$\large{\boxed{a+b+c+d+e+f=12}}$

Tanishq Varshney - 6 years, 1 month ago

Thanks:D, and nicely done! @Tanishq Varshney

Samarpit Swain - 6 years, 1 month ago