If $ I = \int_{0}^{1}\left[1-(1-x^2)^{100}\right]^{201} xdx$ and $J=\int_{0}^{1}\left[1-(1-x^2)^{100}\right]^{202}xdx\;,$ Then $ \frac{I}{J}$

Question

If $\displaystyle I = \int_{0}^{1}\left[1-(1-x^2)^{100}\right]^{201}\cdot xdx$ and $J=\int_{0}^{1}\left[1-(1-x^2)^{100}\right]^{202}\cdot xdx\;,$ Then value of $\displaystyle \frac{I}{J}$

$\bf{My\; Try::}$ Given $$\displaystyle J=\int_{0}^{1}\left[1-(1-x^2)^{100}\right]^{202}\cdot xdx\;,$$

Now Let $(1-x^2) = t\;,$ Then $2xdx = dt$ and Changing Limit

We get $$\displaystyle I = -\frac{1}{2}\int_{1}^{0}(1-t^{100})^{201}dt = \frac{1}{2}\int_{0}^{1}(1-t^{100})^{201}dt$$

And we get $$\displaystyle J = -\frac{1}{2}\int_{1}^{0}(1-t^{100})^{202}dt = \frac{1}{2}\int_{0}^{1}(1-t^{100})^{202}\cdot 1dt$$

Using Integration by parts, We get

$$\displaystyle J = \frac{1}{2}\left[(1-t^{100})^{202}\cdot t\right]_{0}^{1}-\frac{202}{2}\int (1-t^{100})^{201}\cdot (-100t^{100})dt = -10100\int_{0}^{1}(1-t^2)^{201}\left[(1-t^{100})-1\right]dt$$

So Integral $$\displaystyle J = -10100\int_{0}^{1}(1-t^{100})^{202}dt+10100\int_{0}^{1}(1-t^{100})^{201}dt$$

So we get $$\displaystyle J=-10100\cdot 2J+202\cdot 2I\Rightarrow 20201J = 404I\Rightarrow \frac{I}{J} = \frac{20201}{20200}$$

My question is can we solve it using $\bf{Trigonometric\; Substution}$

Or any $\bf{Other\; method,}$ If yes then plz explain here.

Thanks

Answer 1

You can solve it by using the Euler beta function: $$ \int_0^1(1-v)^{a-1}v^{b-1}dv=\frac{\Gamma\left(a \right)\Gamma\left(b \right)}{\Gamma\left(a+b\right)}. $$ You have already proved that $$ I=\int_{0}^{1}\left[1-(1-x^2)^{100}\right]^{201}\cdot xdx=\frac{1}{2}\int_0^1(1-t^{100})^{201}dt, $$ then, performing the change of variable $v=t^{100}$, $dt=\dfrac1{100}v^{1/100-1}dv$, we obtain $$ \begin{align} I&=\frac1{200}\int_0^1(1-v)^{202-1}v^{1/100-1}dv\\\\ I&=\frac1{200}\frac{\Gamma\left(202 \right)\Gamma\left(1/100 \right)}{\Gamma\left(202+1/100\right)}. \end{align} $$ Similarly, $$ \begin{align} J&=\frac1{200}\int_0^1(1-v)^{203-1}v^{1/100-1}dv\\\\ J&=\frac1{200}\frac{\Gamma\left(203 \right)\Gamma\left(1/100 \right)}{\Gamma\left(203+1/100\right)}. \end{align} $$ By using, the property: $$ \Gamma(x+1)=x \: \Gamma(x) $$ we get that $$ \begin{align} \frac{I}{J}&=\frac1{200}\frac{\Gamma\left(202 \right)\Gamma\left(1/100 \right)}{\Gamma\left(202+1/100\right)} \times \frac{200}{1} \frac{\Gamma\left(203+1/100\right)}{\Gamma\left(203 \right)\Gamma\left(1/100 \right)}\\\\ &=\frac1{202}\left(202+\frac1{100}\right) \end{align} $$ giving

$$ \frac{I}{J}=\frac{20201}{20200} $$