And you said Calculus is tough?

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Given that, ∫ d t 2 a t − t 2 = a x sin ⁡ − 1 [ t a − 1 ] \int \frac{dt}{\sqrt{2at-t^{2}}}= a^{x} \sin^{-1} \begin{bmatrix} \frac{t}{a}-1 \end{bmatrix} . The value of x x is?

-1 2 0 1

Vatsalya Tandon by Vatsalya Tandon , 21, India

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1 solution

Tom Engelsman
Oct 11, 2020

If we differentiate both sides with respect to t t , the result becomes:

1 2 a t − t 2 = a x ⋅ ( 1 a ) ( 1 1 − ( t / a − 1 ) 2 ) \frac{1}{\sqrt{2at-t^2}} = a^{x} \cdot (\frac{1}{a})(\frac{1}{\sqrt{1 - (t/a - 1)^2}}) ;

or 1 2 a t − t 2 = a x − 1 ⋅ ( a a 2 − ( t 2 − 2 a t + a 2 ) ) \frac{1}{\sqrt{2at-t^2}} = a^{x-1} \cdot (\frac{a}{\sqrt{a^2- (t^2 -2at +a^2)}}) ;

or 1 2 a t − t 2 = a x ⋅ ( 1 2 a t − t 2 ) \frac{1}{\sqrt{2at-t^2}} = a^{x} \cdot (\frac{1}{\sqrt{2at - t^2 }}) ;

or 1 = a x 1 = a^x ;

or x = 0 . \boxed{x=0}.

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