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Proof by mathematical induction

Int dx / √2 a...

Question

Int dx/√2ax-x,^2 = a^n sin^-1[x/a-1] .

The value of n is
(a) 0 (b) —1
(c) 1 (d) none of these.
You may use dimensional analysis to solve the problem.

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Solution

In $fraction numerator L over denominator square root of L squared minus L squared end root end fraction equals L to the power of 0$ it is speaking about dimension. I will explain it below.

2 numbers can be substracted only if they have same dimension.

that is [2ax]= []

If we take [x]= L, then

Since [2ax]= which gives [a]=[x]=L

Also on substracting 2 quantities having same dimension we get a quantity that is having the same dimension; which gives

$fraction numerator L over denominator square root of L squared minus L squared end root end fraction equals fraction numerator L over denominator square root of L squared end root end fraction equals L over L equals L to the power of 1 cross times L to the power of negative 1 end exponent equals L to the power of 1 minus 1 end exponent equals L to the power of 0$

Now comparing with dimension on RHS
as it represents an angle

Therefore should have dimension

that is

Now in second method how $integral fraction numerator 1 over denominator square root of a squared minus open parentheses x minus a close parentheses squared end root end fraction d x space equals sin to the power of negative 1 end exponent open parentheses fraction numerator x minus a over denominator a end fraction close parentheses$
It is a mathematical derivation which i will show below

$L H S comma space integral fraction numerator 1 over denominator square root of a squared minus open parentheses x minus a close parentheses squared end root end fraction d x t a k e space u space equals fraction numerator x minus a over denominator a end fraction d u space equals 1 over a d x space space space rightwards double arrow d x space equals a d u$

substituting
$integral fraction numerator 1 over denominator square root of a squared minus u squared a squared end root end fraction a d u space equals integral fraction numerator a d u over denominator a square root of 1 minus u squared end root end fraction space space space space space space equals integral fraction numerator 1 over denominator square root of 1 minus u squared end root end fraction d u$

We have a standard integral form as $integral fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction d x equals sin to the power of negative 1 end exponent open parentheses x over a close parentheses$

Applying we get

$integral fraction numerator 1 over denominator square root of 1 minus u squared end root end fraction d u space equals sin to the power of negative 1 end exponent open parentheses u over 1 close parentheses$

substituting for u in terms of x $open parentheses u equals fraction numerator x minus a over denominator a end fraction close parentheses$ we get
$integral fraction numerator 1 over denominator square root of a squared minus open parentheses x minus a close parentheses squared end root end fraction d x equals sin to the power of negative 1 end exponent open parentheses fraction numerator x minus a over denominator a end fraction close parentheses$

It is not neccessary for this derivation by knowing the standard form we can write the integral in single step.

Hope it clears both your doubts and the solution

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Similar questions

\int \frac{d x}{\sqrt{2 a x - x^{2}}} = a^{n} \sin^{- 1} [\frac{x}{a} - 1]

.

The value of n is
(a) 0
(b) −1
(c) 1
(d) none of these.
You may use dimensional analysis to solve the problem.

Q. dx/√2ax-x^2=a^nsin^-1[x/a-1].The value of n is?you may use dimensional analysis to solve the problem.

And:- the value of n is 0 but how ?

\int \frac{x d x}{\sqrt{2 a x - x^{2}}} = a^{n} {sin}^{- 1} [\frac{x}{a} - 1]

. The value of n is:you may use dimensional analysis to solve the problem

Q. You may not know integration, but using dimensional analysis you can check on some results. In the original

\int \frac{d x}{{(2 a x - x^{2})}^{1 / 2}} = a^{n} s i n^{- 1} (\frac{x}{a} - 1)

, find the value of

n

.

\int \frac{d x}{\sqrt{2 a x - x^{2}}} = a^{n} {s i n}^{- 1} [\frac{x - a}{a}]

a =

constant. Using dimensional analysis, the value of

n

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