Verified answer

∫[(sin(2t) + 1)/sin(2t) ]dt

=>∫[sin(2t)/sin(2t)]dt + ∫ dt/2sint cost

=>∫dt + (1/2)∫dt/sint cost

multiply second integral with sec^2(t)/sec^2(t)

=>∫dt + (1/2)∫ [sec^2(t) /(sint)(cost)(sec^2(t)] dt

=>∫dt + (1/2)∫ [sec^2(t)/(sint)cos(t)(1/cos ^2(t)] dt

=>∫dt + (1/2)∫ [sec^2(t)/sint/cos(t)] dt

=>∫dt + (1/2)∫ [sec^2(t)/tant] dt

let tan t = u

sec^2(t) dt = du

=>∫dt + (1/2)∫ du/u

t + (1/2) ln(u)

t + (1/2)ln(tant)

applying limits

[pi/4 - pi/2] + (1/2) [ln(tan(pi/4) - ln(tan(pi/2))]

-pi/4 + (1/2) 0 - undefined]

-pi/4

Which ok do you think of would be maximum useful? we would desire to take the sq. root of (a million-4y^2), so if we make ok =0.5, which will become a million-(cos x)^2 = (sin x)^2, and we are able to take the sq. root. So, enable y = 0.5 cos x. Then dy/dx = -0.5 sin x, so dy = -0.5 sin x dx. The y^2 on the precise turns into 0.25(cos x)^2. finally, what are the obstacles of the essential? because of the fact that we took the valuable sq. root in the previous, we'd prefer to make confident sin x is >= 0. So y=0 while cos x = 0, which provides x = pi/2. y = 0.5 while cos x = a million, so x = 0. So we've the essential from pi/2 to 0 of 0.25(cos x)^2/sin x * -0.5sin x dx = essential from pi/2 to 0 of -0.one hundred twenty five(cos x)^2 dx. to try this, we turn (cos x)^2 into (cos 2x + a million)/2. So we'd prefer to combine -a million/sixteen (cos 2x + a million). That turns into -a million/sixteen (a million/2 sin 2x + x). Substituting in the two limits provides 0 - -pi/32 = pi/32. So its pi/32. i'm ninety 9% confident I made a mistake someplace, yet you get the assumption.