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Ask iit jee aieee pet cbse icse state board experts Expert Question: INTEGRAL CALCULUS........
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[Post New]26 Apr 2007 13:23:11 IST
    Subject: INTEGRAL CALCULUS........
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nayan_dbb (10)

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How do I solve diz integral:
 
cos 8 x dx
 
the answer is :
 
(35x)/128 + (7 sin 2x)/32  + (7sin 4x)/128 +   (sin 6x)/96  +   (sin 8x) / 1024
 
 
 
    
[Post New]26 Apr 2007 18:58:14 IST
    Subject: Re:INTEGRAL CALCULUS........
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vibhu (2)

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put
cos8x=cosx (1-sin2x)9/2
 
then put sinx = t
as cosx is the derivative
 
 
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[Post New]26 Apr 2007 19:14:53 IST
    Subject: Re:INTEGRAL CALCULUS........
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BuddyGuy (83)

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I'm still nt getting it!
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[Post New]26 Apr 2007 19:27:47 IST
    Subject: Re:INTEGRAL CALCULUS........
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aditya_arora04 (1077)

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Hi,
 
Vibhu, your first step may be correct but solving ( 1 - t2 )9/2 would be a herculean task.
 
According to me, first step :  (cos2x)4
Second Step : ( 1 + cos2x )4 / 16 
Open it up.
Then, in the terms where 'cos' occurs in a power-2 or 4, convert it as i said in Step I.
And, then integrate it. It is a bit lengthy, but you will get the answer. And, this question does not come in entrance exams. And has a very less chance of being given in CBSE. But yes, you can do it to strengthen your concepts.
 
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[Post New]26 Apr 2007 19:29:37 IST
    Subject: Re:INTEGRAL CALCULUS........
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un.niranjan (65)

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good work,aditya_arora04.
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[Post New]1 May 2007 20:00:27 IST
    Subject: Re:INTEGRAL CALCULUS........
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avinash.sharma (1189)

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Hello nayan_dbb
 
You can use reduction formula to solve the queries of these types but another method for solving the queries Sinnx  or  Cosnx is 
(i)                if n is odd
 
Step
òSinnx dx
òCosnx dx
 
I
òSin(n-1)x Sinx dx
òCos(n-1)Cosx dx
 
II
ò (1-Cos2x)(n-1)/2 sinx dx
ò (1-Cos2x)(n-1)/2 sinx  dx
cos2x +sin2x =1
III
ò (1-t2) (n-1)/2 dt
 
Sinx =t so cos dx= dt
-ò (1-t2) (n-1)/2 dt
 
Cosx =t so -Sinx dx=dt
Sinx dx =-dt
As n id odd so n-1 is even so (n-1)/2 is an integer
 

IV
Expend these using binomial theorem and integrate.
 
(ii)              if n is even
 
Step
òSinnx dx
òCosnx dx
 
I
ò (Sin2x) n/2 dx
ò(cos2x)n/2 dx
 
II
ò{(1-Cos2x)/2} n/2 dx
ò {(1+Cos2x)/2}n/2 dx
Cos2x = (1+Cos2x)/2
Sin2x = (1-Cos2x)/2
III
(1/2n/2)ò (1-Cos2x) n/2 dt
 
 
(1/2n/2)ò (1+Cos2x) n/2 dt
 
 
As n id even so n/2 is an integer
 

IV
Expend these using binomial theorem and integrate.
In higher order as in the query n=8 you have to apply the both the above process many times. But it is easiest way to solve it.
These rules can also be applied on Sinmx.Cosnx.
 
Now your query is ò Cos8x dx here n is even so
 
= ò (Cos2x)4 dx
 
= ò {(1+Cos2x)/2}4 dx
 
= (1/16) ò (1+Cos2x)4 dx
 
= (1/16) ò (1+ 4Cos2x+6Cos22x+4Cos32x+Cos42x) dx
 
= (1/16) ò (1+ 4Cos2x+6Cos22x+4Cos32x+Cos42x) dx
 
= (1/16) [ò dx +4 òCos2x dx +6 ò Cos22x dx+4ò Cos32x dx+ ò Cos42x dx]
 
= (1/16) [ò dx +4 òCos2x dx +6 ò (1+Cos4x)/2 dx +4 ò (1-Sin22x) Cos2x dx + ò {(1+Cos4x)/2}2 dx]
 
 For underlined quantity use sin2x = t so 2.cos2x dx = dt or cos2x dx =dt/2
 
= (1/16) [ò dx +4 òCos2x dx +3 ò (1+Cos4x) dx +2 ò (1-t2) dt +(1/4) ò (1+Cos4x)2 dx]
 
=(1/16) [ò dx +4 òCos2x dx +3 ò dx+3 ò Cos4x dx +2 ò dt -2 ò t2 dt +(1/4) ò (1+2Cos4x+ Cos24x) dx]
 
 =(1/16) [4ò dx +4 òCos2x dx +3 ò Cos4x dx +2 ò dt -2 ò t2 dt + (1/4)òdx +(1/2)òCos4x dx +(1/4)ò (1+Cos8x)/2 dx]
 
=(1/16) [4ò dx +4 òCos2x dx +3 ò Cos4x dx +2 ò dt -2 ò t2 dt + (1/4) òdx +(1/2)òCos4x dx +(1/8)ò dx +(1/8)òCos8x dx]
 
= (1/16) [ 4x +(4/2) Sin2x +(3/4) Sin4x dx+ 2t ? (2/3) t3 + (1/4) x + (1/8) Sin4x +(1/8)x +(1/64) Sin8x]
 
= (1/16) [ (35/8)x +2Sin2x +(7/8) Sin4x +2Sin2x -(2/3)(Sin2x)3 +(1/64) Sin8x]
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