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Integral Calculus

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Joined:	21 Dec 2007
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6 Jun 2008 12:23:14 IST

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Indefinite Integration Question

None

Find $\int \ (sinx/sin4x) dx$

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C!-!!NMAY

Blazing goIITian

Joined: 30 Mar 2008

Posts: 1415

6 Jun 2008 12:27:53 IST

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sinx/sin4x =sinx/2sin2xcos2x

=sinx/4sinxcosxcos2x

=1/4cosxcos2x

=1/4[secxsec2x]

actually this question is already posted by karan....

and even i m unable to solve further so some body plzz solve it,,,full solution

Marshall Eriksen

Blazing goIITian

Joined: 24 Dec 2006

Posts: 1312

6 Jun 2008 12:44:48 IST

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or simply write sin x as sin (4x - 3x)
expand to get
sin 4x cos 3x - cos 4x sin 3x / sin 4x

cos 3x - sin 3x cot 4x

this is simple to integrate

C!-!!NMAY

Blazing goIITian

Joined: 30 Mar 2008

Posts: 1415

6 Jun 2008 16:24:21 IST

Like 4 people liked this

now finally i got the answer,,,,

so just follow my above sol. up to 1/4 $\int$ [1/cosxcos2x]dx

now multiply with cosx in num and den. both then u will get 1/4 $\int$ [cosx/cos^2xcos2xdx

= >1/4 $\int$ cosx/(1-sin^2x)(1-2sin^2x)dx

=>now put sinx=t and cosxdx=dt,,,,so u will get

I= 1/4 $\int$ dt/(1-t^2)(1-2t^2)..........................(1)

now let t^2=v

then u will get the standard form of 1/(1-v)(1-2v)......

now solve by partial fraction....

plzzz rate me,,,

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