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Integral Calculus

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24 Feb 2007 18:37:26 IST

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a question on integral

None

integration of x⁵/^{[ ]}

x²+1 dx

integration of sin(log x) dx =f(x){sin g(x)- cos h(x)} +c

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CyBorG

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Joined: 6 Jan 2007

Posts: 706

24 Feb 2007 19:35:19 IST

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Here is the solution.

x⁵dx/(

x²+1)=

x⁴xdx/(

x²+1)

Put

1+x²=t or x²=t²-1

xdx/ (

1+x²)=dt

x⁵dx/(

x²+1)=

(t²-1)²dt=

(t⁴-2t²+1)dt=t⁵/5-2t³/3+t-c

Substitute t=

1+x² to get the final answer.

CyBorG

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Joined: 6 Jan 2007

Posts: 706

24 Feb 2007 19:39:52 IST

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Solution for second question.

Let I =

sin(logx)dx=

1.sin(logx)dx=xsin(logx)-

(cos(logx)dx

or I = xsin(logx)-[xcos(logx)+

sin(logx)dx]

or I = xsin(logx)-xcos(logx)-I

or 2I = x(sin(logx)-cos(logx)

or I = (x/2)[sin(logx)-cos(logx)]+c

So f(x)=x/2 and g(x)=h(x)=logx

Bipin Dubey

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Joined: 23 Jan 2007

Posts: 7942

25 Feb 2007 14:02:23 IST

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Substitution : z =

(x²+1)
=> x² = z² - 1
Hence 2x.dx = 2z.dz => x.dx = z.dz

Now I =

x⁵/

(x²+1).dx

=> I =

[(x²)²/

(x²+1)].(x.dx)

=> I =

[(z²-1)²/z](z.dz)

=> I =

[(z²-1)² dz

=> I =

(z⁴-2z²+1).dz = (z⁵/5) - (2z³/3) + z + c

Back substitution : z =

(x²+1)

Gives I = [(x²+1)^5/2/5] - [2(x²+1)^3/2/3] + (x²+1)^1/2 + c

Please post your other query on a new page as we are instructed to answer one query at a time.Hope you understand.

Best Wishes

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