Mathematics - Definite Integration Question with Solution | TestHub

Given,

$f_n={\int }_0^{\pi /2}\left(\sum _{k=1}^n{\sin }^{k-1}x\right)\left(\sum _{k=1}^n(2k-1){\sin }^{k-1}x\right)\cos xdx$

Now let, $\sin x=t$

$\Rightarrow \cos xdx=dt$

So, $f_n={\int }_0^1\left(\sum _{k=1}^n(t{)}^{k-1}\right)\left(\sum _{k=1}^n(2k-1)(t{)}^{k-1}\right)dt$

$\Rightarrow f_n={\int }_0^1\left(1+t+t^2....+t^{n-1}\right)\left(1+3t+5t^2+.....+\left(2n-1\right)t^{n-1}\right)dt$

Now multiply and divide by $\sqrt{t}$ we get,

$\Rightarrow f_n={\int }_0^1\frac{\left(t^{{\displaystyle \frac{1}{2}}}+t^{{\displaystyle \frac{3}{2}}}+t^{\frac{5}{2}}....+t^{{\displaystyle \frac{2n-1}{2}}}\right)}{\sqrt{t}}\left(1+3t+5t^2+.....+\left(2n-1\right)t^{n-1}\right)dt$

$\Rightarrow f_n={\int }_0^1\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}+t^{\frac{5}{2}}....+t^{\frac{2n-1}{2}}\right)\left(t^{\frac{-1}{2}}+3t^{\frac{1}{2}}+5t^{\frac{3}{2}}+.....+\left(2n-1\right)t^{\frac{2n-3}{2}}\right)dt$

Now let $t^{\frac{1}{2}}+t^{\frac{3}{2}}+t^{\frac{5}{2}}....+t^{\frac{2n-1}{2}}=z$

$\Rightarrow \frac{1}{2}\left(t^{\frac{-1}{2}}+3t^{\frac{1}{2}}+5t^{\frac{3}{2}}+.....+\left(2n-1\right)t^{\frac{2n-3}{2}}\right)dt=dz$

Hence, the integral becomes,

$\Rightarrow f_n=2{\int }_0^{n}zdz$

$\Rightarrow f_n={\left[z^2\right]}_0^n=n^2$

Hence, $f_{21}-f_{20}={21}^2-{20}^2=41$