Question Number 129158 by gowsalya last updated on 13/Jan/21
Answered by MJS_new last updated on 13/Jan/21
![∫_0 ^(2π) ∣cos θ +i sin θ −1∣dθ= =2(√2)∫_0 ^π (√(1−cos θ)) dθ= [t=tan (θ/2) → dθ=((2dt)/(t^2 +1))] =8∫_0 ^∞ (t/((t^2 +1)^(3/2) ))dt=−8[(1/( (√(t^2 +1))))]_0 ^∞ =8](Q129190.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\mid\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\:−\mathrm{1}\mid{d}\theta= \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\sqrt{\mathrm{1}−\mathrm{cos}\:\theta}\:{d}\theta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:\rightarrow\:{d}\theta=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\mathrm{8}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{dt}=−\mathrm{8}\left[\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right]_{\mathrm{0}} ^{\infty} =\mathrm{8} \\ $$
Commented by gowsalya last updated on 24/Jan/21
Can u elaborate step by step..ans is correct