If int frac{sin x}{sin ^{3} x+cos ^{3} x} d x | vedclass

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(D) Divide the numerator and denominator by $\cos^3 x$:$I = \int \frac{	an x \sec^2 x}{1+	an^3 x} dx$Let $	an x = t$,then $\sec^2 x dx = dt$:$I = \

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$3$

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(D) Divide the numerator and denominator by $\cos^3 x$:$I = \int \frac{	an x \sec^2 x}{1+	an^3 x} dx$Let $	an x = t$,then $\sec^2 x dx = dt$:$I = \int \frac{t}{(t+1)(t^2-t+1)} dt$Using partial fractions: $\frac{t}{(t+1)(t^2-t+1)} = \frac{A}{t+1} + \frac{Bt+C}{t^2-t+1}$$t = A(t^2-t+1) + (Bt+C)(t+1)$Setting $t = -1$: $-1 = A(1+1+1) \Rightarrow A = -\frac{1}{3}$Comparing coefficients of $t^2$: $A+B = 0 \Rightarrow B = \frac{1}{3}$Comparing constant terms: $A+C = 0 \Rightarrow C = \frac{1}{3}$So,$I = -\frac{1}{3} \int \frac{dt}{t+1} + \int \frac{\frac{1}{3}t + \frac{1}{3}}{t^2-t+1} dt$$I = -\frac{1}{3} \ln|t+1| + \frac{1}{6} \int \frac{2t-1+3}{t^2-t+1} dt$$I = -\frac{1}{3} \ln|t+1| + \frac{1}{6} \ln|t^2-t+1| + \frac{1}{2} \int \frac{dt}{(t-1/2)^2 + 3/4}$$I = -\frac{1}{3} \ln|1+	an x| + \frac{1}{6} \ln|1-	an x+	an^2 x| + \frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{2	an x-1}{\sqrt{3}}ight) + C$Thus,$\alpha = -\frac{1}{3}, \beta = \frac{1}{6}, \gamma = \frac{1}{\sqrt{3}}$$18(\alpha+\beta+\gamma^2) = 18(-\frac{1}{3} + \frac{1}{6} + \frac{1}{3}) = 18(\frac{1}{6}) = 3$.

If $\int \frac{\sin x}{\sin ^{3} x+\cos ^{3} x} d x = \alpha \log _{e}|1+\tan x|+\beta \log _{e}\left|1-\tan x+\tan ^{2} x\right|+\gamma \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C$,where $C$ is the constant of integration,then the value of $18(\alpha+\beta+\gamma^{2})$ is .... .

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