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int(0)^(pi/6) sin2x . cosx dx...

`int_(0)^(pi/6) sin2x . cosx dx `

Text Solution

Verified by Experts

The correct Answer is:
`(2)/(3)-(sqrt3)/(4)`
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Show that (i) int_(0)^(pi//2)f(sinx) d x=int_(0)^(pi//2)f(cos x) d x (ii) int_(0)^(pi//2)f(tan x) d x=int_(0)^(pi//2)f(cot x) d x (iii) int_(0)^(pi//2)f(sin 2 x) sin xd x = int_(o)^(pi//2)f(sin 2x).cosx d x

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Let l= int_(0)^(pi//2)(cosx)/(a cos x+ b sinx)dx and J=int_(0)^(pi//2)(sin x)/(a cosx+b sin x )dx , where a gt 0 and b gt 0 Compute the values of l and j