Question

Let $f\left(x\right)=x\cos x;x\in [\frac{3\pi }{2},2\pi ]$ and $g\left(x\right)$ be its inverse. If ${\int }_0^{2\pi }g\left(x\right)dx=\alpha x^2+\beta \pi +\gamma$, where $\alpha ,\beta$ and $\gamma \in R$ then find the value of $2(\alpha +\beta +\gamma )$.

Answer 1

${\int }_0^{2\pi }g\left(x\right)dx={\int }_{\frac{3\pi }{2}}^{2\pi }f\left(x\right)dx$

Let $u=x$ and $dv=cosxdx⟹v=sinx$. Using chain rule of integration ($\int udv=uv-\int vdu$),

$\int f\left(x\right)dx=xsinx-\int sinx\times 1dx$

$\int f\left(x\right)dx=xsinx+cosx+c$

On applying limits, we have

${\int }_0^{2\pi }g\left(x\right)dx=1+\frac{3\pi }{2}$

$\therefore \alpha =0,\beta =\frac{3}{2}$ and $\gamma =1$

$⟹2(\alpha +\beta +\gamma )=3+2=5$