AP EAMCET · Maths · Indefinite Integration
If \(\int \sqrt{\frac{x}{a^3-x^3}} d x=g(x)+c\), then \(g(x)\) is equal to :
- A \(\frac{2}{3} \cos ^{-1} x\)
- B \(\frac{2}{3} \sin ^{-1}\left(\frac{x^3}{a^3}\right)\)
- C \(\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)\)
- D \(\frac{2}{3} \cos ^{-1}\left(\frac{x}{a}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \sqrt{\frac{x}{a^3-x^3}} d x\) \(=\int \sqrt{\frac{a^3}{a^3-x^3}} \cdot \sqrt{\frac{a^3}{x^3}} \cdot \frac{x^3}{a^3} d x\) \(=\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)+c\) But \(\quad I=g(x)+c\)…
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