Question

please answer this question as soon as possible

Answer 1

[tex]I= \int\ \frac{ x^{2} }{ (x.sinx+cosx)^{2} } dx \\ =\int\ \frac{ x^{2}( sin^{2}x+cos^{2}x ) }{ (x.sinx+cosx)^{2} }dx\\ =\int\ \frac{ x^{2}.sin^{2}x+x^{2}.cos^{2}x }{ (x.sinx+cosx)^{2} }dx\\=\int\ \frac{ x^{2}.sin^{2}x+ x.sinx.cosx-x.sinx.cosx+x^{2}.cos^{2}x }{ (x.sinx+cosx)^{2} }dx\\=\int\ \frac{ x.sinx(x.sinx+cosx)-x.cosx(sinx-x.cosx) }{ (x.sinx+cosx)^{2} }dx\\ u= sinx-x.cosx =\ \textgreater \ du/dx = cosx - 1.cosx +x.sinx = x.sinx \\ v = x.sinx+cox =\ \textgreater \ dv/dx = 1.sinx+x.cosx-sinx=x.cosx\\ I= \int\ {d(u/v)} \,dx\\ [/tex]
I = u/v
I = (sinx-x.cosx)/(x.sinx+cosx)

Answer 2

Answer:

• Sinx - cosx.

Thanks....