NIMCET 2020 — Mathematics PYQ

Concept:

Integration by Parts:
$\int$ f( x)  g( x)  dx= f( x) $\int$g( x)  dx- \int \left [ f( x) \int g( x)  dx\right ]dx

Calculation:

Let us first evaluate $\int ({\mathrm{a}}^2-{\mathrm{x}}^2{)}^{\mathrm{n}}$ dx by integrating it in parts. Let's say f(x)=(a${}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}$ and g(x)=1.

$\therefore I=\int \left[({\mathrm{a}}^2-{\mathrm{x}}^2{)}^{\mathrm{n}}\times 1\right]$dx

$\Rightarrow$I$=({\mathrm{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}\int 1$ dx-$\int [$n$({\mathrm{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}-1}\times 2\mathbf{x}\int 1$ dx]dx

$\Rightarrow$I=x(a${}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}-2$n$\int {\mathbf{x}}^2({\mathbf{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}-1}$d$\mathbf{x}$

$\Rightarrow$I=x(a${}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}+2$n$\int ($a${}^2-{\mathbf{x}}^2-{\mathbf{a}}^2)({\mathbf{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}-1}$d$\mathbf{x}$

$\Rightarrow$I$=\mathbf{x}({\mathbf{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}+2\mathbf{n}[\int ({\mathbf{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}$ $\mathbf{dx}-{\mathbf{a}}^2\int ({\mathbf{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}-1}$ $\mathbf{dx}]$

Since I= $\int$( $a^2$- $x^2{)}^n$ dx,  we get: 

$\Rightarrow$I=x(a${}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}+2$nI-2na${}^2\int ($a${}^2-{\mathbf{x}}^2{)}^{\mathrm{n}-1}$d$\mathbf{x}$

$\Rightarrow (1-2$n)I=x$({\mathrm{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}}-2$na${}^2\int ({\mathrm{a}}^2-{\mathbf{x}}^2{)}^{\mathrm{n}-1}$d$\mathbf{x}$

$\Rightarrow$I$=\frac{\mathrm{x}({\mathrm{a}}^2-{\mathrm{x}}^2{)}^{\mathrm{n}}}{1-2\mathrm{n}}+\frac{2{\mathrm{na}}^2}{2\mathrm{n}-1}\int ({\mathrm{a}}^2-{\mathrm{x}}^2{)}^{\mathrm{n}-2}$

$\Rightarrow I_n$= $\left(\frac{2{\mathrm{na}}^2}{2\mathrm{n}-1}\right)I_n-1$