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प्रश्न
If f(x) = `(2 - xcosx)/(2 + xcosx)` and g(x) = logex, (x > 0) then the value of the integral `int_((-π)/4)^(π/4) "g"("f"(x))"d"x` is ______.
विकल्प
loge1
loge2
logee
loge3
उत्तर
If f(x) = `(2 - xcosx)/(2 + xcosx)` and g(x) = logex, (x > 0) then the value of the integral `int_((-π)/4)^(π/4) "g"("f"(x))"d"x` is `underlinebb(log_e1)`.
Explanation:
f(x) = `(2 - xcosx)/(2 + xcosx)` `{:("g"(x) = log_"e"x),("h"(x > 0)):}`
`int_((-π)/4)^(π/4) "g"("f"(x))"d"x`
`int_((-π)/4)^(π/4) log_"e"((2 - xcosx)/(2 + xcosx))"d"x`
Let h(x) = `log_"e" (2 - xcosx)/(2 + x cosx)`
h(–x) = `log_"e" (2 + xcosx)/(2 - x cosx)`
= `log_"e"((2 - xcosx)/(2 + xcosx))^-1`
= `-log_"e" (2 - xcosx)/(2 + xcosx)` = –f(x)
Odd function
For odd function
`int_((-π)/4)^(π/4)"h"(x)` = 0 = loge1