MT2-2 a) \( \int \frac{1}{\sqrt{9-25 x^{2}}} d x \) AmDCrath Answer any 1 questions from 5 to 6 . Each carries 4 Marks. 5 a) Find the area enclosed by the circle \( x^{2}+y^{2}=25 \) using integration. b) If \( \int \frac{f(x)}{\tan x} d x=\log |\tan x|+C \), then \( f(x)=\ldots \). 6. a) Find the area of the region bounded by \( y^{2}=9 x, x=2, x=4 \) and the \( x \)-axis in the first quadrant. b) Find \( \int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x \) a) Find \( \mathrm{y}^{2}=9 \mathrm{x} n \) - (3) (1) b) \( \int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x \) किणनाइक Answer any 2 questions from 7 to 9 . Each carries 6 Marks . 7. a) The length \( x \) of a rectangle is increasing at the rate of \( 4 \mathrm{~cm} / \mathrm{s} \) and the width \( y \) is decreasing at the rate of \( 5 \mathrm{~cm} / \mathrm{s} \). When \( \mathrm{x}=10 \mathrm{~cm} \) and \( \mathrm{y}=5 \mathrm{~cm} \). i) Find the rates of change of its area. ii) Find the rates of change of the perimeter of the rectangle. \( (2 \) \( (2 \)

To evaluate the integral: \[ \int \frac{1}{\sqrt{9 - 25x^{2}}} \, dx \] we can use a substitution method involving inverse trigonometric functions. ### Step 1: Recognize the Integral Form The integral resembles the standard form: \[ \int \frac{1}{\sqrt{a^{2} - b^{2}x^{2}}} \, dx = \frac{1}{b} \sin^{-1}\left(\frac{b x}{a}\right) + C \] where \( a = 3 \) and \( b = 5 \). ### Step 2: Apply Substitution Let’s set: \[ u = \frac{5x}{3} \quad \Rightarrow \quad x = \frac{3u}{5} \quad \Rightarrow \quad dx = \frac{3}{5} du \] Substituting these into the integral: \[ \int \frac{1}{\sqrt{9 - 25x^{2}}} \, dx = \int \frac{1}{\sqrt{9 - 25\left(\frac{3u}{5}\right)^{2}}} \cdot \frac{3}{5} \, du \] Simplify inside the square root: \[ \sqrt{9 - 25 \cdot \frac{9u^{2}}{25}} = \sqrt{9 - 9u^{2}} = 3\sqrt{1 - u^{2}} \] So the integral becomes: \[ \int \frac{1}{3\sqrt{1 - u^{2}}} \cdot \frac{3}{5} \, du = \frac{1}{5} \int \frac{1}{\sqrt{1 - u^{2}}} \, du \] ### Step 3: Integrate The integral of \( \frac{1}{\sqrt{1 - u^{2}}} \) is \( \sin^{-1}(u) \): \[ \frac{1}{5} \sin^{-1}(u) + C \] ### Step 4: Substitute Back Recall that \( u = \frac{5x}{3} \): \[ \frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C \] ### Final Answer \[ \int \frac{1}{\sqrt{9 - 25x^{2}}} \, dx = \frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C \] Where \( C \) is the constant of integration.