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# C4 integration by parts HELP! watch

1. HELP!

Question: ∫x^2*2sec^2x * tanx dx

I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

I then made V tan^2 X. The answer and solution made V sec^2X.

Is my answer correct. The working in the solution is show below.

----

(e) u=x2 ⇒ du
dx
=2x
dv
dx
=2secxsecxtanx ⇒ v=sec2x
∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx
Let J=∫2xsec2xdx
u=2x ⇒ du
dx
=2
dv
dx
=sec2x ⇒ v=tanx
∴J=2xtanx−∫2tanxdx
=2xtanx−2ln|secx|+ C
∴I=x2sec2x−2xtanx+2ln|secx|+ C′
2. Are you saying the integral of tanx is ln|secx| ?
3. (Original post by thelion0)
HELP!

Question: ∫x^2*2sec^2x * tanx dx

I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

I then made V tan^2 X. The answer and solution made V sec^2X.

Is my answer correct. The working in the solution is show below.

----

(e) u=x2 ⇒ du
dx
=2x
dv
dx
=2secxsecxtanx ⇒ v=sec2x
∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx
Let J=∫2xsec2xdx
u=2x ⇒ du
dx
=2
dv
dx
=sec2x ⇒ v=tanx
∴J=2xtanx−∫2tanxdx
=2xtanx−2ln|secx|+ C
∴I=x2sec2x−2xtanx+2ln|secx|+ C′
We last had this question yesterday.
4. (Original post by m4ths/maths247)
Are you saying the integral of tanx is ln|secx| ?
Are you saying it isn't?
5. (Original post by thelion0)
HELP!

Question: ∫x^2*2sec^2x * tanx dx

I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

I then made V tan^2 X. The answer and solution made V sec^2X.

Is my answer correct. The working in the solution is show below.

----

(e) u=x2 ⇒ du
dx
=2x
dv
dx
=2secxsecxtanx ⇒ v=sec2x
∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx
Let J=∫2xsec2xdx
u=2x ⇒ du
dx
=2
dv
dx
=sec2x ⇒ v=tanx
∴J=2xtanx−∫2tanxdx
=2xtanx−2ln|secx|+ C
∴I=x2sec2x−2xtanx+2ln|secx|+ C′
Not entirely sure why you negged me for referring you to a very recent post that confirmed your approach but your answer is correct so long as you know the difference between and . Your answer is ambiguous in this regard.
6. (Original post by thelion0)
HELP!

Question: ∫x^2*2sec^2x * tanx dx

I took U to be X^2 and I took dv/dx to be 2sec^2x * tanx

I then made V tan^2 X. The answer and solution made V sec^2X.

Is my answer correct. The working in the solution is show below.

----

(e) u=x2 ⇒ du
dx
=2x
dv
dx
=2secxsecxtanx ⇒ v=sec2x
from dv/dx above does not follow v=sec2x

∴I=∫x2×2sec2xtanxdx=x2sec2x −∫2xsec2xdx

As this last integral is x^2 so

same as you wrote correctly, but the way leading to this was correct partly
Let J=∫2xsec2xdx
u=2x ⇒ du
dx
=2
dv
dx
=sec2x ⇒ v=tanx
∴J=2xtanx−∫2tanxdx
=2xtanx−2ln|secx|+ C
∴I=x2sec2x−2xtanx+2ln|secx|+ C′
7. (Original post by Mr M)
Are you saying it isn't?
No of course not, Im just struggling to follow it on a mobile screen given the format.
8. (Original post by m4ths/maths247)
No of course not, Im just struggling to follow it on a mobile screen given the format.
Ah I see. I began to doubt my sanity when I read your previous comment.
9. (Original post by Mr M)
Ah I see. I began to doubt my sanity when I read your previous comment.
No
I was just reading a string of text on a 3" screen thinking of some half decent maths to keep me entertained whilst sitting in a supermarket cafe

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