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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 75 of 160

Marks: +1, -0

If

\int \frac{dx}{(x^2+9)\sqrt{x^2+16}} = \frac{1}{3\sqrt{7}} \tan^{-1}\left(K \frac{x}{\sqrt{16+x^2}}\right) + c

K=

$\frac{\sqrt{7}}{3}$
$3\sqrt{7}$
$\frac{3}{\sqrt{7}}$
$\frac{3}{7}$

Solution:

\text{ Let } I = \int \frac{dx}{(x^2+9)\sqrt{x^2+16}}

\text{ put } x = 4\tan\theta \Rightarrow dx = 4\sec^2\theta\,d\theta

\therefore I = \int \frac{4\sec^2\theta\,d\theta}{(16\tan^2\theta+9)\sqrt{16\tan^2\theta+16}}

= \int \frac{4\sec^2\theta\,d\theta}{(16\tan^2\theta+9)\,4\sec\theta}

= \int \frac{\sec\theta\,d\theta}{16\tan^2\theta+9}

= \int \frac{\cos\theta\,d\theta}{16\sin^2\theta+9\cos^2\theta} = \int \frac{\cos\theta\,d\theta}{7\sin^2\theta+9}

Again put

\sin\theta = t

\Rightarrow \cos\theta\,d\theta = dt

\therefore I = \int \frac{dt}{7t^2+9}

= \frac{1}{7} \cdot \frac{1}{\frac{3}{\sqrt{7}}} \tan^{-1}\left( \frac{t}{\frac{3}{\sqrt{7}}} \right) + C

= \frac{1}{3\sqrt{7}} \tan^{-1}\left( \frac{\sqrt{7}\,t}{3} \right) + C

= \frac{1}{3\sqrt{7}} \tan^{-1}\left( \frac{\sqrt{7}\,\sin\theta}{3} \right) + C

= \frac{1}{3\sqrt{7}} \tan^{-1}\left( \frac{\sqrt{7}\,x}{3\sqrt{x^2+16}} \right) + C

\left(\because x=4\tan\theta \Rightarrow \sin\theta = \frac{x}{\sqrt{x^2+16}}\right)

\therefore K = \frac{\sqrt{7}}{3}

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