Correct option is DWe have∫0π3sinx(2−cosx)2 dxPut 2−cosx=t=>sinx dx=dtIf x=0, then t=1If x=π3, then t=32∫132dtt2=[−1t]132=−23+1=13\begin{aligned}&\text{We have} \\&\int_{0}^{\frac{\pi}{3}} \frac{\sin x}{(2 - \cos x)^2} \, dx \\&\text{Put } 2 - \cos x = t \Rightarrow \sin x \, dx = dt \\&\text{If } x = 0, \text{ then } t = 1 \\&\text{If } x = \frac{\pi}{3}, \text{ then } t = \frac{3}{2} \\&\int_{1}^{\frac{3}{2}} \frac{dt}{t^2} = \left[-\frac{1}{t}\right]_1^{\frac{3}{2}} = -\frac{2}{3} + 1 = \frac{1}{3}\end{aligned}We have∫03π(2−cosx)2sinxdxPut 2−cosx=t=>sinxdx=dtIf x=0, then t=1If x=3π, then t=23∫123t2dt=[−t1]123=−32+1=31
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