Anyone for intergration........??

Question

Need to evaluate the following integrals, please could you show working out and explanations as well, because i need to learn it. Thank you so much.

a) intergrate sec(2x)dx (pi/4
and

b) intergrate 1/(2-t) dt (t>2)

Answer 1

Presumably you've already gone through explanations of how to re-create the first dozen or so entries in a standard table of integrals (see citation).

Here's how to apply them with a simple change of variable.

(a)
|--: Integral from ( π/4 < x <3π/4 ) of sec(2x)dx
If you substitute y=2x (thus dy=2dx), you have the following:
|--: Integral from ( π/2 < y <3π/y ) of 2*sec(y)dy
Hint: The formula for integral of sec(y) dy = ln|sec y + tan y|

(b)
|--: Integral from ( 2 < t < ∞ ) of dt/(2-t)
If you substitute u=2-t (thus du=-t), you have the following:
|--: Integral from ( -∞ < u < 0) of -du/u
Hint: The formula for integral of du/u = ln|u|

Answer 2

a) you'll need to use some trig identities:
sec(2x) = 1/cos(2x)
= cos(2x)/cos^2(2x)
= cos(2x)/(1 - sin^2(2x))

now make the substitution u = sin(2x), so
du = 2cos(2x)

so the integral becomes
1/2 [ du/(1-u^2) ].

then you can use partial fractions to rewrite this as

1/4 [ 1/(1+u) + 1/(1-u) ] du,

which you should then be able to integrate (similarly to part b)

b) let v = 2 - t. then dv = -dt, and the integral is just -dv/v, which is
-ln|v|. don't forget to substitute back in, so you get

-ln|2-t| + C.

Answer 3

a) 0.5*ln | sec 2x + tan 2x | + C
where C is the constant of integration

b) -ln|2-t| + C

Answer 4

wow, not a clue
sorry
x

Answer 5

no