IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Anurag.Seth

Q) Integrate:

(sinx . sin2x . sin3x)dx

abhinav_myself

oh,Thats a easy sum,look

_{[ ]}

^{[ ]} sinx.sin2x.sin3x dx = 1/2 _{[ ]}^{[ ]} 2sinx.sin3x . sin2x

=

_{^1/2[ ]}

^{[ ]} (cos4x-cos2x).sin2x= 1/2_{[ ]}

^{[ ]} cos4x.sin2x- 1/4_{[ ]}

^{[ ]} sin4x

now 1/2

_{[ ]}

^{[ ]} cos4x.sin2x=1/4_{[ ]}

^{[ ]} 2sin2x(1-2sin²x)

Use this ,then use sin³x=3sinx-sin3x/4

Hope u get it

dada

so abhinav

i c that u have started rating ur replies by creating another id(anuhav)....

it is really unfortunate to c u doing something so low as this...

abhinav_myself

hay,dude

sorry guy u misunderstand me .it is my friend anubhav studying in MAV.it is not my id.say your id,i ll send scanned school id of BOTH to your email.

before saying something,u should try to understand other consequences.

AND we r not goin to get any admission in any institute through the points of goiit.also IIT is on its high,i have no time to waste giving unuseful points to eachother.

edison

Use the identity 2sinA sinB = cos(A-B) + cos(A+B)

raman_1290

I =

(sinx.sin2x.sin3x)dx

=

[(sinx)(2sinx.cosx)(3sinx - 4sin³x)]dx

=

6sin³x.cosx.dx-

8sin⁵x.cosx.dx

Put sinx = t ; dt = cosx.dx

=

6t³ -

8t⁵ = (3/2) t⁴ - (4/3) t⁶ + c

Hope its correct.

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