Question

need help the attachment @Calculus1

Answer 1

Didn't I just answer this for you?

Answer 2

yes, but I don't get the last method you show me did you use U-substitution?

Answer 3

We got to the fact that the indefinite integral of this expression is
\[ -\frac{\sec^2 x}{3} - 2\tan x \]

Yes. I didn't fill in the details ,because I was hoping you'd try and see how I did that. The way I did was by substituting u = cos x into
\[ \int \frac{\sin x}{\cos^3 x} \ dx \]

Answer 4

Also you say I needed to integrate them separately, but not together, do mean splitting both expression up. If so do I need to change the order of the endpoint pi/4 and 0

Answer 5

-7/3

Answer 6

but the answer choices for this problem are the following,
a)-2
b)-5/2
c)-1
d)-3/2
e)-1/2

Answer 7

sorry, yes, integral of sin x / cos^3 x dx = -1/2 . sec^2 x
not -1/3

Answer 8

why is -1/2 instead of -1/3

Answer 9

shouldn't that be -u^(-3)

Answer 10

\[ \int u^n \ du = \frac{1}{n+1} u^{n+1} \]
Here n = -3

Answer 11

Oh yes, I dropped a minus sign.

Answer 12

why

Answer 13

I'm having a bad computational day. The indefinite integral is

(1/2) sec^2 x - 2 tan x

Do you agree with that?

Answer 14

yes

Answer 15

Hence evaluating a pi/4 and 0

[ (1/2) 2 - 2 ] - [ 1/2 - 0 ]
= -1 - 1/2
= -3/2

Answer 16

ok

Answer 17

Thanks! for the explanations