Question

Integration of x^2/(xsinx+cosx)^2 dx

Answer 1

I=∫x2(xsinx+cosx)2dx

Apply the Pythagorean trigonometric identity 1=sin2x+cos2x:

I=∫(x2sin2x+x2cos2x(xsinx+cosx)2)dx

Add 0=xsinxcosx−xsinxcosx to the numerator:

I=∫(x2sin2x+xsinxcosx−xsinxcosx+x2cos2x(xsinx+cosx)2)dx

Factorize:

I=∫((xsinx+cosx)(xsinx)+(xcosx−sinx)(xcosx)(xsinx+cosx)2)dx

For ease of understanding, define u(x)=xcosx−sinx and v(x)=xsinx+cosx:

I=∫(vdudx+udvdxv2)dx

This fits the form of the quotient rule, hence:

I=xcosx−sinxxsinx+cosx+C

Answer 2

x^2/(xsinx+cosx)^2 dx
I hope it will help u
Please mark me as Brainliest
App's Name- Photomaths