Can anyone help me to integrate this?

Question

int[e^logx + sinx] cosx dx

answer is xsinx+cosx+ 1/2sin^2x

Answer 1

First of  all e^ logx can be written as x

So now the question becomes

∫ [x + sinx] cosx dx

= ∫ (xcosx + sinxcosx) dx

Now this can be written as separate integrals

= ∫ x cosx dx + ∫ sinx cosx dx

Let us take ∫ xcosx dx and integrate

Use integration by parts

  ÃƒÂ¢Ã‚ˆÂ« x cosx dx = x sinx - ∫ sinx dx

                   = x sinx - (-cos x )

                  = x sinx + cosx...................(1)

Now let us evaluate ∫ sinx cosx dx



Use integration by substitution

        take z = sinx

               dz = cosx dx

∫ sinx cosx = ∫ z dz

                = z²/2

                =(1/2)sin²x...........................(2...

From (1) and (2)

we get the answer

∫ [ e^logx + sinx] cosx dx = x sinx + cosx + (1/2) sin²x

Answer 2

assuming that you mean lnx,

∫ (e^lnx + sinx) cosx dx =

∫ (x + sinx) cosx dx =

expand it into:

∫ (x cosx + sinx cosx) dx =

break it up into:

∫ x cosx dx + ∫ sinx cosx dx =

the first one has to be integrated by parts, assuming:

x = u → dx = du

cosx dx = dv → sinx = v

then integrate, yielding:

∫ u dv = u v - ∫ v du →

∫ x cosx dx + ∫ sinx cosx dx = xsinx - ∫ sinx dx + ∫ sinx cosx dx =

xsinx - (- cosx) + ∫ sinx cosx dx =

xsinx + cosx + ∫ sinx cosx dx =

as for the remaining integral, since it includes both the function sinx and

its derivative cosx, let:

sinx = u

differentiate both sides:

d(sinx) = du →

cosx dx = du

thus, substituting, you get:

xsinx + cosx + ∫ sinx cosx dx = xsinx + cosx + ∫ u du =

xsinx + cosx + [1/(1+1)] u^(1+1) + C =

xsinx + cosx + (1/2) u² + C

that is, being u = sinx:

∫ (e^lnx + sinx) cosx dx = xsinx + cosx + (1/2)sin²x + C

I hope it helps...

Bye!