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Question
`int ("d"x)/(sinx cosx + 2cos^2x)` = ______.
Options
`log |cos x + 2| + "c"`
`log |sin x + 2| + "c"`
`log |tan x + 2| + "c"`
`log |tanx - 2| + "c"`
MCQ
Fill in the Blanks
Solution
`int ("d"x)/(sinx cosx + 2cos^2x)` = `log |tan x + 2| + "c"`.
Explanation:
Let I = `int ("d"x)/(sinx cosx + 2cos^2x)`
= `int (sec^2x)/(tan x + 2) "d"x`
Put tan x + 2 = t
⇒ sec2x dx = dt
∴ I = `int "dt"/"t" = log|"t"| + "c"`
= `log|tanx + 2| + "c"`
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