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Indefinite Integration

Section: Mathematics

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Question : 3 of 89

Marks: +1, -0

If

\int\frac{dx}{2\cos x+3\sin x+4}=\frac{2}{\sqrt{3}}f(x)+c

f\left(\frac{2\pi}{3}\right)=

[AP EAPCET 23 May 2025 S2]

$\frac{\pi}{12}$
$\frac{\pi}{8}$
$\frac{5\pi}{12}$
$\frac{5\pi}{8}$

Solution:

\int\frac{dx}{2\cos x+3\sin x+4}

\text{ Let } t=\tan\frac{x}{2} \Rightarrow \sin x=\frac{2t}{1+t^2}

\cos x=\frac{1-t^2}{1+t^2}

\text{ and } dx=\frac{2dt}{1+t^2}

=\int\frac{2dt}{(1+t^2)\left[\frac{2(1-t^2)}{1+t^2}+\frac{6t}{1+t^2}+4\right]}

=\int\frac{2dt}{2-2t^2+6t+4+4t^2} dt

=\int\frac{dt}{t^2+3t+3}=\int\frac{dt}{\left(t+\frac{3}{2}\right)^2+\frac{3}{4}}

\text{ put } u=t+\frac{3}{2} \Rightarrow du=dt

=\int\frac{1}{u^2+\left(\frac{\sqrt{3}}{2}\right)^2} du

=\frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\left(\frac{u}{\frac{\sqrt{3}}{2}}\right)+C

=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2u}{\sqrt{3}}\right)+C

=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2t+3}{\sqrt{3}}\right)+C

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

\text{ Thus, } f(x)=\tan^{-1}\left(\frac{2\tan\left(\frac{x}{2}\right)+3}{\sqrt{3}}\right)

\therefore f\left(\frac{2\pi}{3}\right)=\tan^{-1}\left(\frac{2\tan\left(\frac{\pi}{3}\right)+3}{\sqrt{3}}\right)

=\tan^{-1}\left(\frac{2\sqrt{3}+3}{\sqrt{3}}\right)

=\tan^{-1}(2+\sqrt{3})=\frac{5\pi}{12}

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