If ∫ sinx sin^3x+ cos^3xdx=α log_e|1+ tanx|+β log_e|1- tanx+ tan^2x|+γ tan^-1( 2 tanx-1 √3)+C, when C is constant of integration, then the value of…

$$\int \frac{\mathrm{sin}x}{{\mathrm{sin}}^{3}x+{\mathrm{cos}}^{3}x}dx=\alpha {\mathrm{log}}_{e}|1+\mathrm{tan}x|+\beta {\mathrm{log}}_{e}|1-\mathrm{tan}x+{\mathrm{tan}}^{2}x|+\gamma {\mathrm{tan}}^{-1}(\frac{2\mathrm{tan}x-1}{\sqrt{3}})+C,..........(i)$$∫sin3x+cos3xsinx​dx=αloge​∣1+tanx∣+βloge​∣1−tanx+tan2x∣+γtan−1(3​2tanx−1​)+C,..........(i)

Let, $\displaystyle I=\int \frac{\mathrm{sin}x}{{\mathrm{sin}}^{3}x+{\mathrm{cos}}^{3}x}dx$I=∫sin3x+cos3xsinx​dx

$$I=\int \frac{\mathrm{tan}x{\mathrm{sec}}^{2}x}{{\mathrm{tan}}^{3}+1}dx$$I=∫tan3+1tanxsec2x​dx

Put $\mathrm{tan}x=t\Rightarrow {\mathrm{sec}}^{2}xdx=dt$tanx=t⇒sec2xdx=dt

$$=\int \frac{tdt}{{t}^{3}+1}=\int \frac{t}{(t+1)({t}^{2}-t+1)}dt$$=∫t3+1tdt​=∫(t+1)(t2−t+1)t​dt

Now $\frac{t}{(t+1)({t}^{2}-t+1)}=\frac{A}{t+1}+\frac{Bt+C}{{t}^{2}-t+1}$(t+1)(t2−t+1)t​=t+1A​+t2−t+1Bt+C​

$$t=A({t}^{2}-t+1)+(Bt+C)(t+1)$$t=A(t2−t+1)+(Bt+C)(t+1)

$$t={t}^{2}(A+B)-t(A-B-C)+A+C$$t=t2(A+B)−t(A−B−C)+A+C

Comparing on both sides

$$A+B=0,-A+B+C=1,A+C=0$$A+B=0,−A+B+C=1,A+C=0

Solving these equations

$$A=-\frac{1}{3},B=\frac{1}{3},C=\frac{1}{3}$$A=−31​,B=31​,C=31​

Hence $\displaystyle I=\int {\frac{-\frac{1}{3}}{t+1}+\frac{1}{3}(\frac{t+1}{{t}^{2}-t+1})}dt$I=∫t+1−31​​+31​(t2−t+1t+1​)dt

$$=-\frac{1}{3}\int \frac{1}{(t+1)}dt+\frac{1}{3}\int \frac{\frac{1}{2}(2t-1)+\frac{3}{2}}{{t}^{2}-t+1}dt$$=−31​∫(t+1)1​dt+31​∫t2−t+121​(2t−1)+23​​dt

$\displaystyle =-\frac{1}{3}\int \frac{1}{(t+1)}dt+\frac{1}{6}\int \frac{2t-1}{({t}^{2}-t+1)}+\frac{1}{2}\int \frac{dt}{{(t-\frac{1}{2})}^{2}+{(\frac{\sqrt{3}}{2})}^{2}}$=−31​∫(t+1)1​dt+61​∫(t2−t+1)2t−1​+21​∫(t−21​)2+(23​​)2dt​$=-\frac{1}{3}\mathrm{ln}|t+1|+\frac{1}{6}\mathrm{ln}|{t}^{2}-t+1|+\frac{1}{2}\cdot \frac{2}{\sqrt{3}}{\mathrm{tan}}^{-1}(\frac{2t-1}{\sqrt{3}})+C$=−31​ln∣t+1∣+61​ln∣t2−t+1∣+21​⋅3​2​tan−1(3​2t−1​)+C

$$=-\frac{1}{3}\mathrm{ln}|\mathrm{tan}x+1|+\frac{1}{6}\mathrm{ln}|{\mathrm{tan}}^{2}x-\mathrm{tan}x+1|+\frac{1}{\sqrt{3}}{\mathrm{tan}}^{-1}(\frac{2\mathrm{tan}x-1}{\sqrt{3}})+C$$=−31​ln∣tanx+1∣+61​ln∣tan2x−tanx+1∣+3​1​tan−1(3​2tanx−1​)+C

From equation $(i)$(i), Comparing from both sides

$\Rightarrow \alpha =-\frac{1}{3},\beta =\frac{1}{6}$⇒α=−31​,β=61​ and $\gamma =\frac{1}{\sqrt{3}}$γ=3​1​

So $18(\alpha +\beta +{\gamma }^{2})$18(α+β+γ2)

$$=18(-\frac{1}{3}+\frac{1}{6}+\frac{1}{3})$$=18(−31​+61​+31​)

$$=3$$=3