Question

∫[(x²dx)/(xsinx + cosx)²] = _______ + c 

(a) [(sinx – xcosx)/(xsinx + cosx)] 

(b) [(sinx + xcosx)/(xsinx + cosx)] 

(c) [(xsinx – cosx)/(xsinx + cosx)] 

(d) [(xsinx + cosx)/(xsinx – cosx)]

Answer 1

The correct option (a) [(sinx – xcosx)/(xsinx + cosx)]   

Explanation:

I = ∫[(x²dx)/(xsinx + cosx)²]dx 

= ∫(xsecx) ∙ [(xcosx)/(xsinx + cosx)²]dx 

= (xsecx) × [(– 1)/(xsinx + cosx)] – ∫(secx + xsecxtanx) × [(– 1)/(xsinx + cosx)]dx 

[As ∫{(xcosx)/(xsinx + cosx)²}dx = ∫(dt/t²) where t = xsinx + cosx] 

= [(– xsecx)/(xsinx + cosx)] + ∫sec²xdx 

= [(– x)/{cosx(xsinx + cosx)}] + tanx + c 

= [{– x + sinx(xsinx + cosx)}/{cosx(xsinx + cosx)}] + c 

= [{– x + x(1 – cos²x) + sinxcosx}/{cosx(xsinx + cosx)}] + c 

= [(– x + x – xcos²x + sinxcosx)/{cosx(xsinx + cosx)}] + c 

= [{cosx(sinx – xcosx)}/{cosx(xsinx + cosx)}] + c 

= [(sinx – xcosx)/(xsinx + cosx)] + c