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Question
`int "e"^sintheta [log(sin theta) + "cosec"^2theta]costheta "d"theta` = ______.
Options
`"e"^sintheta [log(sin theta) + "cosec"^2theta] + "c"`
`"e"^sintheta [log(sin theta) + "cosec"theta] + "c"`
`"e"^sintheta [log(sin theta) - "cosec"theta] + "c"`
`"e"^sintheta [log(sin theta) - "cosec"^2theta] + "c"`
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Solution
`int "e"^sintheta [log(sin theta) + "cosec"^2theta]costheta "d"theta` = `"e"^sintheta [log(sin theta) - "cosec"theta] + "c"`.
Explanation:
Let I = `int "e"^sintheta [log(sin theta) + "cosec"^2theta]costheta "d"theta`
Put sin θ = t
⇒ cos θ dθ = dt
∴ = `int "e"^"t"[log"t" + 1/"t"^2]"dt"`
= `int"e"^"t" [log"t" 1/"t" + 1/"t" + 1/"t"^2] "dt"`
= `"e"^"t" (log "t" - 1/"t") + "c"` .....`[because "d"/"dt" (log "t" - 1/"t") = 1/"t" + 1/"t"^2]`
= `"e"^sintheta[log(sin theta) - 1/sintheta] + "c"`
= `"e"^sintheta [log(sin theta) - "cosec" theta] + "c"`
