Question

Integration as soon as possible

Answer 1

Let, I=∫x2(xsinx+cosx)2dx,

=∫{(xsecx)(xcosx(xsinx+cosx)2)}dx.

We will use the following Rule of Integration by Parts (IBP) :

IBP : ∫uv'dx=uv−∫u'vdx.

Prior to the Integration, let us note :

ddx{1xsinx+cosx}

=−1(xsinx+cosx)2⋅ddx{(xsinx+cosx)},

=−1(xsinx+cosx)2⋅{(x⋅cosx+sinx)+(−sinx)}.

=−xcosx(xsinx+cosx)2.

⇒∫xcosx(xsinx+cosx)2dx=−1xsinx+cosx............(1).

Also, ddx(xsecx)=xsecxtanx+secx,

=x⋅1cosx⋅sinxcosx+1cosx.

⇒ddx(xsecx)=xsinx+cosxcos2x............(2).

Now, in IBP, we take,

u=xsecx,and,v'=xcosx(xsinx+cosx)2.

∴u'=xsinx+cosxcos2x,&,v=−1xsinx+cosx.

∴I=(xsecx){−1xsinx+cosx}

−∫{(xsinx+cosxcos2x)⋅(−1xsinx+cosx)}dx,

=−xcosx(xsinx+cosx)+∫sec2xdx,

=−xcosx(xsinx+cosx)+tanx,

=−xcosx(xsinx+cosx)+sinxcosx,

=−x+sinx(xsinx+cosx)cosx(xsinx+cosx),

=−x+xsin2x+sinxcosxcosx(xsinx+cosx),

=−x(1−sin2x)+sinxcosxcosx(xsinx+cosx),

=−xcos2x+sinxcosxcosx(xsinx+cosx),

=cosx(sinx−xcosx)cosx(xsinx+cosx).

⇒I=sinx−xcosxxsinx+cosx+C.

Hope it helps