Question Number 106869 by bemath last updated on 07/Aug/20
![@bemath@ lim_(x→π) [((1/(2x−2π))) ∫_π ^x ((cos 2t dt)/(1−cos 3t)) ]=?](Q106869.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left[\left(\frac{\mathrm{1}}{\mathrm{2x}−\mathrm{2}\pi}\right)\:\underset{\pi} {\overset{\mathrm{x}} {\int}}\:\frac{\mathrm{cos}\:\mathrm{2t}\:\mathrm{dt}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3t}}\:\right]=? \\ $$
Answered by bobhans last updated on 07/Aug/20
![^(∧bobhans∧) lim_(x→π) [((∫_π ^x ((cos 2t dt)/(1−cos 3t)))/(2x−2π)) ] = lim_(x→π) (((d/dx) ∫^x ((cos 2t dt)/(1−cos 3t)))/((d/dx)[2x−2π])) = lim_(x→π) (((((cos 2x)/(1−cos 3x))))/2) = ((1/(1+1))/2) = (1/4)](Q106874.png)
$$\:\:\:\:\:\:\:\:\:\overset{\wedge\mathrm{bobhans}\wedge} {\:} \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left[\frac{\underset{\pi} {\overset{\mathrm{x}} {\int}}\:\frac{\mathrm{cos}\:\mathrm{2t}\:\mathrm{dt}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3t}}}{\mathrm{2x}−\mathrm{2}\pi}\:\right]\:=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\frac{\mathrm{d}}{\mathrm{dx}}\:\overset{\mathrm{x}} {\int}\:\frac{\mathrm{cos}\:\mathrm{2t}\:\mathrm{dt}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3t}}}{\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{2x}−\mathrm{2}\pi\right]} \\ $$$$=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{cos}\:\mathrm{2x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3x}}\right)}{\mathrm{2}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$