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Question: $\int \frac{sin x}{3+4cos^2x}dx = Atan^{-1}(Bcosx)+c$ where c is a constant of integration. Then, t...

∫sinx3+4cos2xdx=Atan−1(Bcosx)+c\int \frac{sin x}{3+4cos^2x}dx = Atan^{-1}(Bcosx)+c

where c is a constant of integration. Then, the value of A+B is:

A

523\frac{5}{2\sqrt{3}}

B

−123\frac{-1}{2\sqrt{3}}

C

−23\frac{-2}{\sqrt{3}}

D

32\frac{\sqrt{3}}{2}

Answer

32\frac{\sqrt{3}}{2}

Explanation

Solution

To evaluate the integral ∫sin⁡x3+4cos⁡2xdx\int \frac{\sin x}{3+4\cos^2x}dx, we use the method of substitution.

Let u=cos⁡xu = \cos x. Differentiating both sides with respect to xx, we get: du=−sin⁡x dxdu = -\sin x \, dx This implies sin⁡x dx=−du\sin x \, dx = -du.

Substitute these into the integral: ∫sin⁡x3+4cos⁡2xdx=∫−du3+4u2\int \frac{\sin x}{3+4\cos^2x}dx = \int \frac{-du}{3+4u^2} =−∫du4u2+3= -\int \frac{du}{4u^2+3} Factor out 4 from the denominator: =−14∫duu2+34= -\frac{1}{4}\int \frac{du}{u^2+\frac{3}{4}} This integral is of the standard form ∫dxx2+a2=1atan⁡−1(xa)+C\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C. In our case, x=ux=u and a2=34a^2 = \frac{3}{4}, so a=34=32a = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.

Applying the formula: −14[132tan⁡−1(u32)]+c-\frac{1}{4}\left[ \frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\left(\frac{u}{\frac{\sqrt{3}}{2}}\right) \right] + c =−14[23tan⁡−1(2u3)]+c= -\frac{1}{4}\left[ \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2u}{\sqrt{3}}\right) \right] + c =−123tan⁡−1(2u3)+c= -\frac{1}{2\sqrt{3}}\tan^{-1}\left(\frac{2u}{\sqrt{3}}\right) + c Now, substitute back u=cos⁡xu = \cos x: =−123tan⁡−1(2cos⁡x3)+c= -\frac{1}{2\sqrt{3}}\tan^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right) + c

The given form of the integral is Atan⁡−1(Bcos⁡x)+cA\tan^{-1}(B\cos x)+c. Comparing our result with the given form: A=−123A = -\frac{1}{2\sqrt{3}} B=23B = \frac{2}{\sqrt{3}}

We need to find the value of A+B: A+B=−123+23A+B = -\frac{1}{2\sqrt{3}} + \frac{2}{\sqrt{3}} To add these fractions, we find a common denominator, which is 232\sqrt{3}: A+B=−123+2×223A+B = -\frac{1}{2\sqrt{3}} + \frac{2 \times 2}{2\sqrt{3}} A+B=−123+423A+B = -\frac{1}{2\sqrt{3}} + \frac{4}{2\sqrt{3}} A+B=−1+423A+B = \frac{-1+4}{2\sqrt{3}} A+B=323A+B = \frac{3}{2\sqrt{3}} To simplify, we can rationalize the denominator by multiplying the numerator and denominator by 3\sqrt{3}: A+B=323×33A+B = \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} A+B=332×3A+B = \frac{3\sqrt{3}}{2 \times 3} A+B=336A+B = \frac{3\sqrt{3}}{6} A+B=32A+B = \frac{\sqrt{3}}{2}

The value of A+B is 32\frac{\sqrt{3}}{2}.