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Integral Calculus

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19 Jan 2007 21:21:32 IST

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SOLVE THIS IF YOU CAN

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

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nibin89

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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.

CyBorG

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20 Jan 2007 08:38:49 IST

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Sorry.The ans is not zero.

nikhil_chandra2008

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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.

CyBorG

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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

malay

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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

CyBorG

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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.

Aditya Arora

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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 ????

CyBorG

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22 Jan 2007 07:27:28 IST

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See the signature part.

CyBorG

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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)

prashanth10000

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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

abhinav_myself

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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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