Question

If $I_t={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}tx}{{\sin }^{2}x}dx$ then ,$I_1,I_2,I_3$ are in

• A. A.P.
• B. H.P.
• C. G.P.
• D. None of these

Answer 1

Answer: (A)

A.P.

Consider $I_{t+2}+I_t-2I_{t+1}$
${\int }_0^{\pi /2}\frac{({\sin }^{2}tx+{\sin }^2(t+2)x-2{\sin }^2(t+1)x)dx}{{\sin }^{2}x}$
${\int }_0^{\pi /2}\frac{({\sin }^2(t+2)x-{\sin }^2(t+1)x-{\sin }^2(t+1)x+{\sin }^{2}tx)dx}{{\sin }^{2}x}$
${\int }_0^{\pi /2}\frac{[\sin (2t+3)x-\sin (2t+1)x]dx}{\sin x}$
$=2{\int }_0^{\pi /2}\cos (2t+2)xdx=2{\left(\frac{\sin (2t+2)x}{2(t+1)}\right)}_0^{\pi /2}$
$=\frac{1}{(t+1)}\left(0\right)=0$
$\therefore I_t,I_{t+1},I_{t+2}ϵA.P.$
Short Cut Method :
Substitute $t=1,2,3$ in given integral we get
$I_1=\frac{\pi }{2},I_2=\pi ,I_3=\frac{3\pi }{2}$
$I_1,I_2,I_3ϵA.P.$