Question

(b) Differentiate the following functions with respect to x: (i) ln(1 + sin2 x) (ii) xx.
(c) Evaluate the integralZ 2x3 −4x−8 x4 −x3 + 4x2 −4x dx.

Answer 1

i) Let $u(x) = 1 + \sin2x.$ Then $\ln(1+\sin 2x) = \ln u(x).$

$\dfrac{d}{dx}(\ln(1+\sin 2x)) = \dfrac{d}{dx}(\ln u(x)) = \dfrac{1}{u(x)}\cdot u'(x)~~~(\text{by chain rule}) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= \dfrac{1}{1+\sin2x}\cdot 2\cos2x\\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= \dfrac{2\cos2x}{1+\sin2x}$

ii) The given function can be written as $x^x=e^{\ln x^{x}} = e^{x\ln x}.$ Let $v(x) = x \ln x.$

Therefore,

$\dfrac{d}{dx}(x^{x}) = \dfrac{d}{dx}(e^{v(x)})\\~\\~~~~~~~~~~~~~~=e^{v(x)}\cdot v'(x)~~~~~(\text{by chain rule})\\ ~~~~~~~~~~~~~~=e^{x \ln x} (x \cdot \frac{1}{x}+\ln x)~~(\text{using product rule})\\~~~~~~~~~~~~~~=x^{x}(1 + \ln x)$

c)$\displaystyle\int\dfrac{2x^{3}-4x-8}{x^{4}-x^{3}+4x^{2}-4x}dx = \displaystyle2\int\dfrac{x^{3}-2x-4}{x(x-1)(x^{2}+4)}dx.$

Using partial fraction for the integrand,

$\dfrac{x^{3}-2x-4}{x(x-1)(x^{2}+4)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{Cx+D}{x^{2}+4}\\~\\ x^{3}-2x-4=A(x-1)(x^{2}+4)+Bx(x^{2}+4)+(Cx+D)(x^{2}-x)$

Solving for A, B, C and D we get

$A = 1, B =-1, C = 1, D = 2$

Thus,

$\displaystyle\int\dfrac{2x^{3}-4x-8}{x^{4}-x^{3}+4x^{2}-4x}dx = \displaystyle 2\int\biggl(\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{x+2}{x^{2}+4}\biggr)dx\\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\displaystyle 2 \biggl(\ln x -\ln (x-1)\biggr) + \int\dfrac{2x}{x^{2}+4}dx+2\int\dfrac{2}{x^{2}+4}dx\\~ \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\ln x - 2\ln (x-1)+ \ln(x^{2}+4)+2\arctan\bigg(\dfrac{x}{2}\bigg)\\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln x^{2}-\ln(x-1)^{2}+\ln(x^{2}+4)+2\arctan\bigg(\dfrac{x}{2}\bigg)\\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln\bigg(\dfrac{x^{2}(x^{2}+4)}{(x-1)^{2}}\bigg)+2\arctan\bigg(\dfrac{x}{2}\bigg).$