Question

If \( \int \frac{dx}{2\cos x + 3\sin x + 4} = \frac{2}{\sqrt{3}}f(x) + c \), then \( f\left(\frac{2\pi}{3}\right) = \)

• A. \( \frac{\pi}{12} \)
• B. \( \frac{\pi}{8} \)
• C. \( \frac{5\pi}{12} \)
• D. \( \frac{5\pi}{8} \)

Hint:

For integrals of the form \( \int \frac{dx}{a\cos x + b\sin x + c} \), the half-angle tangent substitution \( t = \tan\left(\frac{x}{2}\right) \) is generally the most effective method. This transforms the trigonometric integral into an algebraic integral involving rational functions of \(t\), which can then be solved using standard integration techniques such as completing the square and partial fractions. Remember the key relations: \( \sin x = \frac{2t}{1+t^2} \), \( \cos x = \frac{1-t^2}{1+t^2} \), and \( dx = \frac{2dt}{1+t^2} \).

Answer 1

The Correct Option is C

Step 1: Use the half-angle substitution to simplify the integrand.
Let \( t = \tan\left(\frac{x}{2}\right) \). The relevant identities are: \[ \sin x = \frac{2t}{1+t^2} \] \[ \cos x = \frac{1-t^2}{1+t^2} \] \[ dx = \frac{2dt}{1+t^2} \] Substitute these into the integral: \[ \int \frac{\frac{2dt}{1+t^2}}{2\left(\frac{1-t^2}{1+t^2}\right) + 3\left(\frac{2t}{1+t^2}\right) + 4} \] Multiply the numerator and denominator by \((1+t^2)\) to clear the denominators: \[ = \int \frac{2dt}{2(1-t^2) + 3(2t) + 4(1+t^2)} \] Expand and combine like terms in the denominator: \[ = \int \frac{2dt}{2 - 2t^2 + 6t + 4 + 4t^2} \] \[ = \int \frac{2dt}{2t^2 + 6t + 6} \] Factor out 2 from the denominator: \[ = \int \frac{dt}{t^2 + 3t + 3} \] Step 2: Complete the square in the denominator and integrate.
Complete the square for the denominator \( t^2 + 3t + 3 \). Half of the coefficient of \(t\) is \( \frac{3}{2} \), and squaring it gives \( \left(\frac{3}{2}\right)^2 = \frac{9}{4} \). \[ t^2 + 3t + 3 = t^2 + 3t + \frac{9}{4} - \frac{9}{4} + 3 \] \[ = \left(t + \frac{3}{2}\right)^2 + \frac{12-9}{4} \] \[ = \left(t + \frac{3}{2}\right)^2 + \frac{3}{4} \] Now the integral becomes: \[ \int \frac{dt}{\left(t + \frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \] This integral is of the form \( \int \frac{du}{u^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C \). Here, \( u = t + \frac{3}{2} \) and \( a = \frac{\sqrt{3}}{2} \). \[ = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{t + \frac{3}{2}}{\frac{\sqrt{3}}{2}}\right) + c \] \[ = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{\frac{2t+3}{2}}{\frac{\sqrt{3}}{2}}\right) + c \] \[ = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t+3}{\sqrt{3}}\right) + c \] Step 3: Relate the result to the given form and find \(f(x)\).
We are given that \( \int \frac{dx}{2\cos x + 3\sin x + 4} = \frac{2}{\sqrt{3}}f(x) + c \).
Comparing our integrated result with the given form: \[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t+3}{\sqrt{3}}\right) + c = \frac{2}{\sqrt{3}}f(x) + c \] Therefore, \( f(x) = \tan^{-1}\left(\frac{2t+3}{\sqrt{3}}\right) \). Substitute back \( t = \tan\left(\frac{x}{2}\right) \): \[ f(x) = \tan^{-1}\left(\frac{2\tan\left(\frac{x}{2}\right)+3}{\sqrt{3}}\right) \] Step 4: Calculate \( f\left(\frac{2\pi}{3}\right) \).
Substitute \( x = \frac{2\pi}{3} \) into the expression for \(f(x)\): \[ f\left(\frac{2\pi}{3}\right) = \tan^{-1}\left(\frac{2\tan\left(\frac{1}{2} \cdot \frac{2\pi}{3}\right)+3}{\sqrt{3}}\right) \] \[ = \tan^{-1}\left(\frac{2\tan\left(\frac{\pi}{3}\right)+3}{\sqrt{3}}\right) \] Since \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \): \[ = \tan^{-1}\left(\frac{2\sqrt{3}+3}{\sqrt{3}}\right) \] Factor out \( \sqrt{3} \) from the numerator: \[ = \tan^{-1}\left(\frac{\sqrt{3}(2+\sqrt{3})}{\sqrt{3}}\right) \] \[ = \tan^{-1}\left(2+\sqrt{3}\right) \] We know that \( \tan\left(\frac{5\pi}{12}\right) = \tan(75^\circ) \).
Using the tangent addition formula: \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Let \( A = 45^\circ \) and \( B = 30^\circ \).
\( \tan(45^\circ+30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \) Rationalize the denominator:
\( \frac{\sqrt{3}+1}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3+1+2\sqrt{3}}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3} \)
Thus, \( \tan^{-1}\left(2+\sqrt{3}\right) = \frac{5\pi}{12} \).