Question

need help on the attachment @Calculus1

Answer 1

(tan x - 2)/cos^2 x = sin x / cos^3 - 2 sec^2 x

Now you should be able to integrate both of those terms separately.

Answer 2

where the sec^(2)x and the cos^(3)

Answer 3

come from?

Answer 4

c'mon.
tan x = sin x / cos x
1 / cos x = sec x

Answer 5

I knew that, but why is cos^(2) and not just cos x and same thing for the sec^(2), why isn't it just 1/sec x?

Answer 6

\[ \frac{\tan x - 2}{\cos^2 x} = \frac{\tan x}{\cos^2 x} - \frac{2}{\cos^2 x} \]
\[ = \frac{\sin x}{\cos x}.\frac{1}{\cos^2} - 2 \left( \frac{1}{\cos x} \right)^2\]
\[ = \frac{\sin x}{\cos^3 x} - 2 (\sec x)^2 \]
\[ = \frac{\sin x}{\cos^3 x} - 2 \ \sec^2 x \]

Answer 7

oh, ok

Answer 8

I got -2 is that correct

Answer 9

I don't think so. What do you have for the indefinite integral?

Answer 10

-1/3ln(u)-2tanx and the endpoint a=0 and b=pi/4

Answer 11

No. This ln u term is wrong. The tan x is right.

Answer 12

u=cos^3(x)
du=-3sinxdx
-1/3du=sinxdx

Answer 13

No.
\[ \int \sin x / \cos^3 x \ dx = \int \sin x (\cos x)^{-3} \ dx = -1/3 . (\cos x)^{-2} = -1/3 . \sec^2 x \ \ \]

Answer 14

did you use u-substitution?