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Integral Calculus

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 Joined: 6 Jan 2007 Post: 706
19 Jan 2007 21:21:32 IST
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Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Integral Calculus

Integral from -(negative sign)pi/3 to pi/3 of ((pi+4x^3)dx)/(2-(cos(|x|+pi/3)))

Cool goIITian

Joined: 17 Jan 2007
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19 Jan 2007 22:43:48 IST
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is the limit from pi/3 to pi/3. r u sure????

then ans is 0.

Forum Expert
Joined: 6 Jan 2007
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20 Jan 2007 08:38:49 IST
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Sorry.The ans is not zero.

New kid on the Block

Joined: 20 Jan 2007
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20 Jan 2007 09:00:04 IST
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hey,man,if the limit is from pi\3 to pi\3 then it has to be zero.plzzz do tell how to solve if this is wrong.

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Joined: 6 Jan 2007
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20 Jan 2007 13:21:32 IST
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See the limit properly. It is from -pi/3 to pi/3

Hot goIITian

Joined: 18 Dec 2006
Posts: 146
20 Jan 2007 16:06:03 IST
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Let denominator be D
break the integrals into two integrals
pi/D + 4x^3/D
the second evalutes to zero( propertiy of an odd function)

break the first integral again in two parts, in first take limit -pi/3 to 0, in secont take limit 0 to pi/3

I hope it helps

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21 Jan 2007 13:43:37 IST
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Malay I know the ans as well as solution.I wanted to know who will solve it.

Blazing goIITian

Joined: 3 Dec 2006
Posts: 1043
21 Jan 2007 22:17:37 IST
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Do it as follows, my friend :

1.  [-pie/3 ][pie/3 ]  (pie + 4x3) dx / 2- cos(|x| + pie/3)

2.[-pie/3 ][pie/3 ] pie dx / 2-cos(|x| + pie/3) + [-pie/3 ][pie/3 ] 4x3 /  2- cos(|x| + pie/3)

3.  Second function is an odd function and hence its integration is 0.
4.  [-pie/3 ][pie/3 ] pie dx / 2-cos(|x| + pie/3)

5.  2 [0][pie/3 ] pie dx / 2-cos(|x| + pie/3)

6.  As limits are 0 to pie/3 that is +ve, so instead of |x| we can write x.
7.   2 [0][pie/3 ] pie dx / 2-cos(x + pie/3)
8.   2 pie [0][pie/3 ] dx / 2-cos(x + pie/3)
9.   Open up cos(x + pie/3) by cos(A + B)=cosAcosB - sinAsinB

10.  Now put sinx=2tan(x/2) / (1+tan2x) and cosx=(1-tan2x) / ( 1+tan2x ) and
then you can solve it easily.

Please do tell me if i am wrong somewhere or if you encounter any problem.

Note : Whats your real name KAB ????

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22 Jan 2007 07:27:28 IST
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See the signature part.

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Joined: 6 Jan 2007
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22 Jan 2007 16:19:06 IST
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It is quite an easy one.I thought it was a good one and I posted it before solving it completely.
Now I hope this is a tough one.
Evaluate
(tan-1x)dx/(1+x)

New kid on the Block

Joined: 7 Feb 2007
Posts: 9
7 Feb 2007 03:19:33 IST
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applying integration by parts ,
tan-1x.ln(1+x) -
[ ][ ] [ln(1+x)]/(1+x2) dx

put x+1 =t in 2nd integral nd again apply integration by parts u will get d answer

sorry if there is any mmistake

New kid on the Block

Joined: 17 Jan 2007
Posts: 18
7 Feb 2007 15:42:38 IST
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sorry v cant substitue t=1+x.then how to convert 1+x2 ?

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