Answer to Question #124209 in Calculus for desmond

Question

Evaluate the following integrals: (i)Z xln(x + 1)dx [13 marks] (ii)Z sin3(lnx)cos2(lnx) x dx

Answer 1

$(i) \int$ xln(x+1)dx

= ln(x+1)$\int$ xdx - $\int [\frac{d}{dx}(ln(x+1)) \int xdx]dx$

= $\frac {x^2ln(x+1)}{2}$ - $\int \frac{x^2}{2(x+1)}$ dx

= $\frac {x^2ln(x+1)}{2}$ - $\int \frac{x^2-1+1}{2(x+1)}dx$

= $\frac {x^2ln(x+1)}{2}$

- $\frac{1}{2}$ $[\int \frac{x^2-1}{(x+1)}dx + \int \frac{1}{(x+1)}dx]$

= $\frac {x^2ln(x+1)}{2}$ - $\frac{1}{2}$ [ $\int$(x-1)dx+ $\int\frac{1}{(x+1)}dx$ ]

= $\frac {x^2ln(x+1)}{2}$ - $\frac{1}{2}$ [ $\frac{x^2}{2}$ - x + ln(x+1)] + C

= $\frac {x^2ln(x+1)}{2}$ - $\frac{1}{2}$ $\frac{x^2-2x}{2}$ - $\frac {ln(x+1)}{2}$ + C

= $\frac {(x^2-1)ln(x+1)}{2} - \frac{x^2-2x}{4} +$ C where C is integration constant.

(ii)

I = $\int$ sin 3(lnx) cos 2(lnx) x dx

So 2I = $\int$ 2 sin 3(lnx) cos 2(lnx) x dx

= $\int$ x[sin 5(lnx) + sin (lnx)]dx

= $\int$ xsin 5(lnx)dx + $\int$xsin (lnx)dx

= I₅ + I₁ (let)

Where I₅ = $\int$ xsin 5(lnx)dx

and I₁ = $\int$ xsin (lnx)dx

Now let us find a general formula for $I_m=$ $\int$ xsin m(lnx)dx

$I_m=\int$ xsin m(lnx)dx

$I_m$ = sin m(lnx) $\int$ xdx $-$

$\int$ $[\frac{d}{dx}(sin m(lnx))\int xdx]dx$

$I_m$ = $\frac{x^2}{2}sin m(lnx)-\int \frac{mcos m(lnx)}{x} \frac {x^2}{2}dx$

$I_m$ = $\frac{x^2}{2}sin m(lnx)-$ $\frac{m}{2}\int xcos m(lnx) dx$

$I_m$ = $\frac{x^2}{2}sin m(lnx)-$ $\frac {m}{2} [cos m(lnx)\int xdx$ - $\int (\frac{d}{dx}(cos m(lnx))\int xdx)]$

$I_m$ = $\frac{x^2}{2}sin m(lnx) -$ $\frac{m}{2}$ [$\frac{x^2}{2}cos m(lnx)-$ $\int - \frac{msin m(lnx)}{x} \frac {x^2}{2}dx]$

$I_m$ = $\frac{x^2}{2}sin m(lnx)$ - $\frac{m}{2}$ [$\frac{x^2}{2}cos m(lnx)$ + $\frac{m}{2}$ $\int$ x sin m(lnx)dx]

$I_m$ = $\frac{x^2}{2}sin m(lnx)$ - $\frac{m}{2}$ $\frac{x^2}{2}cos m(lnx)$ - $\frac{m^2}{4}$ $I_m$

So $(1+\frac{m^2}{4})$ $I_m$ = $\frac{x^2}{4}[2sin m(lnx)-mcos m(lnx)]$

So $I_m$ = $\frac{x^2}{m^2+4}[2sin m(lnx)-mcos m(lnx)]$

So putting m = 5 we get

I₅= $\frac{x^2}{5^2+4}[2sin 5(lnx)-5cos 5(lnx)]$

= $\frac{x^2}{29}[2sin 5(lnx)-5cos 5(lnx)]$

and putting m = 1 we get

I₁= $\frac{x^2}{1^2+4}[2sin (lnx)-cos (lnx)]$

= $\frac{x^2}{5}[2sin (lnx)-cos (lnx)]$

So 2I = $\frac{x^2}{29}[2sin 5(lnx)-5cos 5(lnx)]$ + $\frac{x^2}{5}[2sin (lnx)-cos (lnx)]$

=> I = $\frac{x^2}{58}[2sin 5(lnx)-5cos 5(lnx)]$ + $\frac{x^2}{10}[2sin (lnx)-cos (lnx)]$ + C

So $\int$ sin 3(lnx) cos 2(lnx) x dx

= $\frac{x^2}{58}[2sin 5(lnx)-5cos 5(lnx)]$ + $\frac{x^2}{10}[2sin (lnx)-cos (lnx)]$ + C

where C is integration constant