Use the substitution $x = 3\sec \theta$ to find \(\displaystyle\int \frac{1}{x^2 \sqrt{x^2 - 9}} \

Question

Use the substitution $x = 3\sec \theta$ to find $\displaystyle\int \frac{1}{x^2 \sqrt{x^2 - 9}} \, \mathrm{d}x$.[5] The region $R$ lies in the first quadrant and is bounded by the curve $y^2 = \frac{1}{x^2 \sqrt{x^2 - 9}}$ and the lines $y = 0.1$ and $x = 4$. Find the exact volume of the solid generated when $R$ is rotated through $2\pi$ radians about the $x$-axis.[4]

Accepted Answer

(a) $\frac{\sqrt{x^2-9}}{9x}+C$, for $x>3$ or $x<-3$. (b) $\pi(\frac{71}{900}-\frac{\sqrt7}{36})$ cubic units.

2025 SAJC P1 Q9

2025 SAJC P1 Q9

Solution:

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