If ∫sin3x+cos3xsinxdx=αloge∣1+tanx∣+βloge∣1−tanx+tan2x∣+γtan−1(32tanx−1)+C, where C is constant of integration, then the value of 18(α+β+γ2) is
Answer: 3
Solutions · 1
Official solution
\[I=\int\frac{\sin x}{\sin^3x+\cos^3x}dx=\int\frac{\tan x\sec^2x}{\tan^3x+1}dx\]
Put tanx=t⇒sec2xdx=dt:
\[I=\int\frac{t\,dt}{t^3+1}=\int\frac{t\,dt}{(t+1)(t^2-t+1)}\]
By partial fractions, (t+1)(t2−t+1)t=t+1A+t2−t+1Bt+C gives A=−31,B=31,C=31. So
\[I=-\frac13\ln(t+1)+\frac16\ln(t^2-t+1)+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+C\]
Comparing: α=−31,β=61,γ=31
\[18\left(\alpha+\beta+\gamma^2\right)=18\left(-\frac13+\frac16+\frac13\right)=18\cdot\frac16=3\]