If In- xn-a2-x2dx and n-k- In-xn-1a2-x2p-n-1a2- In-2- then 3k-2p-where m-n-m-n-2

Given :(n+k)· I_n= x^n 1(a^2 x^2)^p+(n 1)a^2· I_n 2 nbsp;⋯(i) and nbsp;I_n=∫ x^n√(a^2 x^2)dx = 12∫ x^n 1( 2x·√(a^2 x^2)) dx Let u=x^n 1, v= 2x·√(a^2 x^2 ...

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Byju's Answer

Standard XII

Mathematics

Iota

If In=∫ xn√a2...

Question

If $I_{n} = \int x^{n} \sqrt{a^{2} - x^{2}} d x$ and $(n + k) \cdot I_{n} = - x^{n - 1} (a^{2} - x^{2})^{p} + (n - 1) a^{2} \cdot I_{n - 2},$ then $3 k - 2 p =$
(where $m, n \in N; m, n \geq 2$ )

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Solution

Given : $(n + k) \cdot I_{n} = - x^{n - 1} (a^{2} - x^{2})^{p} + (n - 1) a^{2} \cdot I_{n - 2} \dots (i)$
and $I_{n} = \int x^{n} \sqrt{a^{2} - x^{2}} d x = - \frac{1}{2} \int x^{n - 1} (- 2 x \cdot \sqrt{a^{2} - x^{2}}) d x$
Let $u = x^{n - 1}, v = - 2 x \cdot \sqrt{a^{2} - x^{2}}$ and applying integration By part
$\Rightarrow - 2 I_{n} = x^{n - 1} \cdot \frac{2}{3} (a^{2} - x^{2})^{\frac{3}{2}} - \int (n - 1) \cdot x^{n - 2} \cdot \frac{2}{3} (a^{2} - x^{2})^{\frac{3}{2}} d x = x^{n - 1} \cdot \frac{2}{3} (a^{2} - x^{2})^{\frac{3}{2}} - \frac{2}{3} \cdot (n - 1) \int x^{n - 2} \cdot (a^{2} - x^{2}) (a^{2} - x^{2})^{\frac{1}{2}} d x = x^{n - 1} \cdot \frac{2}{3} (a^{2} - x^{2})^{\frac{3}{2}} - \frac{2}{3} \cdot (n - 1) \int (a^{2} \cdot x^{n - 2} - x^{n}) (a^{2} - x^{2})^{\frac{1}{2}} d x = x^{n - 1} \cdot \frac{2}{3} (a^{2} - x^{2})^{\frac{3}{2}} - \frac{2}{3} \cdot (n - 1) [a^{2} \cdot I_{n - 2} - I_{n}]$
$\Rightarrow - 3 I_{n} - (n - 1) I_{n} = x^{n - 1} (a^{2} - x^{2})^{\frac{3}{2}} - (n - 1) \cdot a^{2} \cdot I_{n - 2}$
$∴ (n + 2) I_{n} = - x^{n - 1} (a^{2} - x^{2})^{\frac{3}{2}} + (n - 1) \cdot a^{2} \cdot I_{n - 2}$
Comparing with $(i)$ we get, $k = 2, p = \frac{3}{2}$
$∴ 3 k - 2 p = 3$

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If In=∫xn√a2−x2dx and (n+k)⋅In=−xn−1(a2−x2)p+(n−1)a2⋅In−2, then 3k−2p= (where m,n∈N;m,n≥2)

If $I_{n} = \int x^{n} \sqrt{a^{2} - x^{2}} d x$ and $(n + k) \cdot I_{n} = - x^{n - 1} (a^{2} - x^{2})^{p} + (n - 1) a^{2} \cdot I_{n - 2},$ then $3 k - 2 p =$
(where $m, n \in N; m, n \geq 2$ )