Answer to Question #104843 in Calculus for SHIVAM KUMAR

Question

Derive the reduction formula
∫ (x^2+a^2)^n\2 dx= x (x^2+a^2)^n\2\n+1+na^2\n+1∫ (x^2+a^2)^n\2-1 dx
Use the formula to integrate ∫ (x^2+a^2)5\2

Answer 1

ANSWER.

To derive the formula

$\int { { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }dx } =\frac { 1 }{ n+1 } x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }+\frac { n }{ n+1 } { a }^{ 2 }\int { { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n-2 }{ 2 } }dx }$ (1)

we use integration by parts $\int { udv } =uv-\int { vdu }$ . We denote

${ I }_{ n\quad }:=\int { { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }dx }$ . Then (1) has the form:

${ I }_{ n\quad }=\frac { 1 }{ n+1 } x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }+\frac { n }{ n+1 } { a }^{ 2 }{ I }_{ n-2\quad }$ (2)

If $u={ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } },\quad dv=dx$ , then $v=x,\quad du=nx{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } -1 }dx$ and

${ I }_{ n\quad }=x\cdot { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }-n\int { { { x }^{ 2 }\left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } -1 }dx } =$

$=x\cdot { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }-n\int { { { (x }^{ 2 }+{ a }^{ 2 }-{ a }^{ 2 })\left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n-2 }{ 2 } \quad }dx } =$

$=x\cdot { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }+n{ a }^{ 2 }{ I }_{ n-2\ }-n{ I }_{ n\quad \quad }$.Add to both sides of the last equality  $(nI_n)$ ,we get $(n+1){ I }_{ n\ }=x\cdot { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { n }{ 2 } }+n{ a }^{ 2 }{ I }_{ n-2\quad }$ . Dividing both sides of the equality by $(n+1)$ , we obtain the formula (2) .

We use formula (2) to calculate the integral  ${ I }_{ 5 }\ =\int { { \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 5 }{ 2 } }dx\ } (n=5)$

$I_5=\frac { 1 }{ 5+1 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 5 }{ 2 } }+\frac { 5{ a }^{ 2 } }{ 5+1 } { I }_{ 3 }\quad\quad (3)$

$I_3=\frac { 1 }{ 3+1 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+\frac { 3{ a }^{ 2 } }{ 3+1 } { I }_{ 1 }\quad\quad (4)$

$I_1=\frac { 1 }{ 1+1 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 1 }{ 2 } }+\frac { \ { a }^{ 2 } }{ 1+1 } { I }_{ -1 }\quad\quad (5)$

${ I }_{ -1 }=\ln { \left| x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right| } +Const$

Substituting (4) in (3), (5) in (4),  $I_{-1}$ in (5) , we obtain

$I_5=\frac { 1 }{ 6 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 5 }{ 2 } }+\frac { 5{ a }^{ 2 } }{ 6 } { I }_{ 3 }=$

$=\frac { 1 }{ 6 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 5 }{ 2 } }+\frac { 5{ a }^{ 2 } }{ 6 } \left[ \frac { 1 }{ 4 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+\frac { 3{ a }^{ 2 } }{ 4 } { I }_{ 1 } \right] =$

$=\frac { 1 }{ 6 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 5 }{ 2 } }+\frac { 5{ a }^{ 2 } }{ 24 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+\frac { 15{ a }^{ 4 } }{ 24 } \left[ \frac { 1 }{ 2 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 1 }{ 2 } }+\frac { \ { a }^{ 2 } }{ 2 } { I }_{ -1 } \right] =$

$=\frac { 1 }{ 6 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 5 }{ 2 } }+\frac { 5{ a }^{ 2 } }{ 24 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+$

$+\frac { 15{ a }^{ 4 } }{ 48 } \cdot x{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ \frac { 1 }{ 2 } }+\frac { 15{ a }^{ 6 } }{ 48 } \cdot \ \ln { \left| x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right| } +C$