Could any one possibly help me these questions. I have written the question and then my attempt at the solution next.

Find each of the following indefinite integrals, identifying any general rules

of calculus that you use.

[integral]xcos(1/3x)dx

Solving the equation by integration by parts. Let f(x)=x and g'(x)=cos(1/3x); then f'(x)=1 and g(x)=3 sin(1/3x) Substituting into the equation for integration by parts f(x)g(x)-∫f ' (x)g(x)dx. we obtain ∫xcos(1/3x)dx=3xsin(1/3x)-∫1(3 sin(1/3)x)=3xsin(1/3x)-3∫sin(1/3)x=3xsin(1/3 x)+9cos(1/3 x)+C

x/√1-x^4

Take u=1-x^4; thendu/dx=-4x^3. Hence∫x/√(1-x^4) dx=x/√u (-4x^3 )dx=x/√u du= -1/2 u^(-3/2). Plugging u back into the equation1/2 4x^3 ^(-3/2)

Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies

below the x-axis for 1 < x < 2 and above the x-axis for 2 < x < 3.

The graph of the function f(x)=(x^2+1) ln(1/2x) is positive above the x-axis and negative below the x-axis.

Use this fact to find the area enclosed by the graph and the x-axis

between x = 1 and x = 3, giving your answer to five decimal

places.

I have no idea how to answer this.

Find the volume of the solid of revolution obtained when the graph of

f(x)= sec x + tan x, from x=−[pie]/3 to x = [pie]/4

is rotated about the x-axis. Give your answer to four decimal places.

The volume is [pie]∫(-[pie]/3) ([pie]/4) secx+tanx=[pie](ln(tan(x)+sec(x) )+ln(sec(x)))=[pie]([(ln(tan(-[pie]/3)+sec(-[pie]/3)+ln(sec(-[pie]/3) ]-[(ln(tan([pie]/4)+sec([pie]/4)+ln(sec([pie]/4)]=[pie [ln(-√3+2)+ln(2)-[(ln(1+√2)+ln(√2)]=[pie][-0.63-1.2)=-1.83[pie]