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Integral Calculus

Blazing goIITian

Joined:	22 Jul 2007
Post:	712

11 Apr 2008 21:36:00 IST

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952

fyjg

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integrate the following(plz give the complete solution)
1)e^2logcotx   ans -cotx-x
2)^{[ ]} (1-x)/(1+x)   ans sin^- x +^{[ ]} (1-x²)
3)cosh2xsinhx   ans (e^3x -e^-3x )/12 - (e^x - e^-x )/4
4)sec⁵xtanx    ans    1/5(sec⁵x)
5)cos³xsinx ans -(cos⁴x )/4
6)^{[ 3]} sinx      ans 3/4(sinx)^4/3
7)x³ sinx⁴   ans -(cosx⁴ )/4
8)cot²xcosecx²x ans -(cot³x)/3
9)(sec²x)/^{[ ]} (1-tan²x)     ans sin^-1 (tanx)
10)1/(1-cosx-sinx) ans logI1-cotx/2I
11)sin²x   ans 1/2{x-1/2sin2x}
12)sin² (ax+b)   ans 1/a{(1/2(ax+b)-1/4sin² (ax+b)}
13)cos3xcosx   ans   (1/10)sin5x +(1/2)sinx
14)cosxcos2xcos3x   ans (1/24)sin6x +(1/16)sin4x +(1/8)sin2x +(1/4)x
15) ^{[ ]} (2+sin3x) cos3x   ans (1/9 )(2(2+sin3x)^3/2 )
16) sin2x/(^{[ ]} 1+cos²x)      ans -2^{[ ]} (1+cos²x)
17)sin2x/(a+bcosx)² ans   -(1/2){2log Ia+bcosxI} - {2a/(b²(a+bcosx))}
many salutes assured

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Comments (14)

Akhil

Blazing goIITian

Joined: 17 Jan 2008

Posts: 1451

11 Apr 2008 21:37:33 IST

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1)
e^{log cot^2 x}=cot²x

_{[ ]}

^{[ ]} cot²xdx=_{[ ]}

^{[ ]} cosec²x-1 dx
=-cot x -x

shrithi

Cool goIITian

Joined: 30 Mar 2008

Posts: 68

11 Apr 2008 21:50:06 IST

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let x=sin

dx/d

=cos

_{[ ]}

^{[ ]}

1-x^2 =_{[ ]}^{[ ]}

1-(sin

)^2 {cos

=_{[ ]}

^{[ ]} cos³

now use reduction formula n integrate this....

shrithi

Cool goIITian

Joined: 30 Mar 2008

Posts: 68

11 Apr 2008 21:55:02 IST

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3) cosh2x=(e^2x+ e^-2x)/2 and sinhx=(e^x - e^-x)/2

substitute these in the ques n ull get the ans

shrithi

Cool goIITian

Joined: 30 Mar 2008

Posts: 68

11 Apr 2008 21:58:43 IST

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let secx=t

(secx)(tanx)dx=dt

thus we get.....

_{[ ]}

^{[ ]} t^4dt =t^5/5

swati purnima

Scorching goIITian

Joined: 3 Dec 2007

Posts: 260

11 Apr 2008 22:17:01 IST

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fr d 5th que here goes d solution

given

_{[ ]}

^{[ ]} cos^3 xsinx dx

den v know dat

let cosx =t den

-sinx dx =dt

substitute d value of cos x & on further simplification u will get d value of

given integral

swati purnima

Scorching goIITian

Joined: 3 Dec 2007

Posts: 260

11 Apr 2008 22:17:36 IST

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rate me if satisfied

swati purnima

Scorching goIITian

Joined: 3 Dec 2007

Posts: 260

11 Apr 2008 22:29:22 IST

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fr d 8th que in applying identies u get
integralcot^4x+cot^2xdx
& on simplification u get d required value

fr d 10th problem cosx=1-tan^2x/1+tan^2x
& similarly fr sinx & den integrate u will get d answer
fr d 11th problem sin^2x=1-cos2x/2

& den simplyfy u will get d answer

rohit

Blazing goIITian

Joined: 24 Jan 2008

Posts: 496

11 Apr 2008 22:30:07 IST

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11)
sin^2x

now cos2x=1- 2sin^2 x
so sin^2= (1-cos2x)/2

now integrate using normal integrations

12) it is also like 11...........continue by same method

rohit

Blazing goIITian

Joined: 24 Jan 2008

Posts: 496

11 Apr 2008 22:36:43 IST

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13) use2 cosAcosB = cos (A+B) + cos (A-B)

in this case we get (cos 4x + cos 2x)/2

then its normal integration

i think the 13 ans is wrong.........it is not matching with the calculated answer

14) it is on the same line..................just take first two terms and find the value in sum form using the same formula
ten u take the third term with these terms

then u get the answer using normal integration

rohit

Blazing goIITian

Joined: 24 Jan 2008

Posts: 496

11 Apr 2008 22:45:05 IST

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15) use sin3x=t
then its normal calculations

16) use cos^2x=t
then this becomes normal calculations

9) use tanx =t

8) use cot x = t

7)x^4 =t

ashish banga

Blazing goIITian

Joined: 16 Dec 2006

Posts: 1362

11 Apr 2008 22:49:08 IST

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9) tanx = t
sec^2xdx = dt .
it is simple afterwards

varsha valli g.

Blazing goIITian

Joined: 31 Jan 2007

Posts: 515

12 Apr 2008 10:30:18 IST

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4)i=secxtanx(sec^4x) put secx=t dt=secxtanx

5)put cosx=t dt=(-sinx)

7)multiply and divide by 4

i=1/4 (4x^3sin(x^4)) put x^4=t dt=4x^3

8)put cotx=t dt=(-cosecx)^2

9)put tanx=t

!!!!!!!!!!!

varsha valli g.

Blazing goIITian

Joined: 31 Jan 2007

Posts: 515

12 Apr 2008 10:36:27 IST

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17)put a+bcosx=1
dt = -bsinx dx
cosx=(t-a)/b

substituting,v get i= (-2/b^2) *(t-a)/t^2

splitting into 2 integrals
i=(-2/b^2)(1/t - a/t^2)

now u can integrate it
i think the first term is a bit wrong in the ans u gave.
!!!!!!!!!!!!!!!!!!

Uday Prakash

Blazing goIITian

Joined: 7 Dec 2006

Posts: 504

16 Apr 2008 23:04:39 IST

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2) Just rationalize, u'll get (1-x)/rt(1-x^2)

Now just split and integrate....

sin^-1x -rt(1-x²)

~Cheerio!!!

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