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Integral Calculus

Shreya's Avatar
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12 Apr 2008 21:29:06 IST
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can u plz integrate these
1)sec5xtanx    ans    1/5(sec5x)
2)(cos2x)/(sinx +cosx)2 =   ans   log lsinx+cosxl
3)(xe-1   +  ex-1 )/(xe   +   ex )  ans 1/e{logIxe +exl}
can u also tell me how(on wat basis)  we should  take substitutions like t,z etc
4)(sin2x)/(sin5xsin3x)  ans 1/3LogIsin3xI-1/5logIsinxI
5)(1-tanx)/(1+tanx)     ans  log Icosx+sinxI
6)cot3x  ans -1/2cot2x - log lsinxl
7)(sin2x)/(sin5xsin3x)   ans   1/3loglsin3xl  - 1/5 loglsin5xl
8)1/(a2sin2x + b2cos2x)  ans 1/ab(tan-1(atanx/b)
9)1/(ex + e-x )2   ans -1/(2(1+e2x ))
10) 1/(ex +1)(1+e-x )   ans -1/(1+ex )
11)(x+2)/ (1-x2 )        ans - (1-x2 )   + 2sin-x
12)cosec4x  ans  -cotx  -(cot3x)/3
13) tan3x   ans  -loglsecxl   +(tan2 x)/2


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Comments (4)

varsha valli g.'s Avatar

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12 Apr 2008 21:44:54 IST
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1)i=sec^4x secxtanx put t=secx dt =secxtanx

2)cos2x=cos^2x - sin^2x
substitute and put t=sinx +cosx dt=cosx-sinx

3)put t=x^e+e^x dt=e(x^e-1+e^x-1)

5)tanx=sinx/cosx substitute,proceed same as 2nd
!!!!!!!!!!!!!!!
varsha valli g.'s Avatar

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12 Apr 2008 21:50:32 IST
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6)cotx * cot^2x=cotx(cosec^2x-1)
put t=cotx

8)take out a^2cos^2x as common from the denom,it will bcom sec^2x *1/a^2 in numerator
i=(1/a^2)sec^2x/(tan^2x+(b^2/a^2))
put t=tan^2x
apply the formula .in the formula A=b/a
varsha valli g.'s Avatar

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12 Apr 2008 21:55:46 IST
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9)e^-x =1/e^x
multiply and divide by 2 ,put t=e^2x +1 dt=2e^2x

10)e^-x=1/e^x
put e^x+1=t dt=e^x

13)i=tanx*tan^2x=tanx(sec^2x-1)
put t=tanx dt=sec^2x

!!!!!!!!!!!!!!!!!!!!!!!!!
Uday Prakash's Avatar

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16 Apr 2008 22:35:04 IST
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4=7)

sin2x=sin(5x-3x)

(sin5xcos3x-cos5xsin3x)sin5xsin3x = cot3x - cot5x

Integrating this, the answer is 1/3ln|sin3x| - 1/5ln|sin5x|


 11)

x/rt(1-x2) + 2/rt(1-x2)

Now integrating this, -rt(1-x2) +2sin-1x

12) csc4x = csc2x(cot2x+1)

               = csc2x + csc2xcot2x

Integrating this u get -cotx -1/3cot3x

~Cheerio!!!



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