evaluate integral with upper limit=pi,lower limit=0; xsinxdx/(1+cos^2x)?

Answer 1

u=x dv=sinx dx/ (1+cos^2 x)
du=dx v=???

z=1+cos^2 x dz=-2 sin x dx
so dv=-1/2 dz/z v=-1/2 ln (1+cos^2 x)

uv-integral of v du = -1/2 x ln (1+cos^2 x) + 1/2 ln(1+cos^2 x) dx

evaluating the first term -1/2[pi ln(1+1)-0] = -pi/2 ln (2)

-pi/2 ln (2) + 1/2 ln(1+cos^2 x) dx.......a little closer

Answer 2

Which ok do you think of would be maximum functional? we'd desire to take the sq. root of (a million-4y^2), so if we make ok =0.5, which will develop right into 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 will become 0.25(cos x)^2. ultimately, what are the bounds of the indispensable? provided that we took the helpful sq. root until now, we'd want to make sure 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 indispensable from pi/2 to 0 of 0.25(cos x)^2/sin x * -0.5sin x dx = indispensable from pi/2 to 0 of -0.a hundred twenty five(cos x)^2 dx. to try this, we turn (cos x)^2 into (cos 2x + a million)/2. So we'd want to combine -a million/sixteen (cos 2x + a million). That will become -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% particular I made a mistake someplace, yet you get the assumption.