IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

shinee

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

akhil_o

1)
e^{log cot^2 x}=cot²x

_{[ ]}

^{[ ]} cot²xdx=_{[ ]}

^{[ ]} cosec²x-1 dx
=-cot x -x

shrithi

let x=sin

dx/d

=cos

_{[ ]}

^{[ ]}

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

1-(sin

)^2 {cos

}d

=_{[ ]}

^{[ ]} cos³

d

now use reduction formula n integrate this....

shrithi

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

let secx=t

(secx)(tanx)dx=dt

thus we get.....

_{[ ]}

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

amulye

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

amulye

rate me if satisfied

amulye

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

dream

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

dream

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

dream

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

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

varshavallig

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

!!!!!!!!!!!

varshavallig

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.
!!!!!!!!!!!!!!!!!!

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