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प्रश्न
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
बेरीज
उत्तर
`int 1/[ 2 - 3cos2x] dx`
As `cos 2x = 2cos^x - 1`
So `int 1/[ 2 - 3cos2x] dx = int 1/[2 - 3( 2cos^x - 1) ]`
And multiply and divide by sec2x
Then we have `int sec^2x/[5sec^2x - 6]` dx
= `int (sec^2x)/[5( 1 + tan^2x) - 6]`
= `int (sec^2x dx)/( 5tan^2x - 1)`
Let tan x = t, then sec2x dx = dt
Hence `int [sec^2x]/[ 5tan^2x - 1]`
= `int dt/[5t^2 - 1]`
= `1/5 int dt/[t^2 - (1/sqrt5)^2]`
= `1/5 log |[ t - 1/sqrt5 ]/[ t + 1/sqrt5 ]|`
= `1/5 log |[ tan x - 1/sqrt5]/[ tan x + 1/sqrt5 ]| + c`
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