Solveeit Logo
Login

Question

Mathematics Question on Methods of Integration

The value of the integral ∫dxxx2−a2\int \frac{dx}{x \sqrt{x^{2} - a^{2}} } is equal to:

A

c+1asin⁡−1a∣x∣c + \frac{1}{a} \sin^{-1} \frac{a}{|x|}

B

c−1asin⁡−1a∣x∣c - \frac{1}{a} \sin^{-1} \frac{a}{|x|}

C

c−1acos⁡−1a∣x∣c - \frac{1}{a} \cos^{-1} \frac{a}{|x|}

D

sin⁡−1a∣x∣+c\sin^{-1} \frac{a}{|x|} + c

Answer

c−1asin⁡−1a∣x∣c - \frac{1}{a} \sin^{-1} \frac{a}{|x|}

Explanation

Solution

Let I=∫dxxx2−a2I=\int \frac{d x}{x \sqrt{x^{2}-a^{2}}}
Let, x=1tx=\frac{1}{t}
∴ dx=−1t2 dt\therefore \, d x =-\frac{1}{t^{2}} \,d t
∴ I=∫−dtt2⋅1t(1t2)2−a2=−1a∫dt(1a)2−t2\therefore\, I =\int \frac{-d t}{t^{2} \cdot \frac{1}{t} \sqrt{\left(\frac{1}{t^{2}}\right)^{2}-a^{2}}}=-\frac{1}{a} \int \frac{d t}{\sqrt{\left(\frac{1}{a}\right)^{2}-t^{2}}}
=−1asin⁡−1t+C=−1asin⁡−1a∣x∣+C=-\frac{1}{a} \sin ^{-1} t+C=-\frac{1}{a} \sin ^{-1} \frac{a}{|x|}+C
=C−1asin⁡−1a∣x∣=C-\frac{1}{a} \sin ^{-1} \frac{a}{|x|}