Evaluate `∫ 3^x/ sqrt(1 − 9^x ) d x `

Answer

`a rarr q; b rarr s; c rarr p; d rarr r`.

Explanation

a. `5 sin theta + 3 (sin theta cos alpha - cos theta sin alpha)`
`=(5+3 cos alpha) sin theta-3 sin alpha cos theta`
`rArr underset(theta in R)("max") {5 sin theta+3 sin (theta - alpha)}`
`=sqrt((5+3 cos alpha)^(2)+9 sin^(2) alpha)`
`=sqrt(34+30 cos alpha)`
Therefore, the given equation is
`34+30 cos alpha=49`
or `cos alpha =1//2`
`rArr alpha =2 n pi pm pi//3, n in Z`
b. `(cos x)^(sin^(2) x-3 sin x+2)=1`
`rArr sin^(2) x-3 sin x+2=0`
`rArr (sin x-2) (sin x-1)=0`
`rArr sin x=1`
But this does not satisfy the equation because `0^(@)=1` is absurd.
c. `sqrt((sin x)) + 2^(1//4) cos x=0` ...(i)
`:. sqrt((sin x)) gt 0` and so `cos x lt 0`,
Also `sin x gt 0 rArr x` lies in the second quadrant
Equation (i) can be rewritten as
`2^(1//4) cos x=-sqrt((sin x))`
On squaring, we get `sqrt(2) cos^(2) x = sin x`
or `sqrt(2) sin^(@) x+sin x-sqrt(2)=0`
or `(sqrt(2) sin x-1) (sin x + sqrt(2))=0`
`sin x ne -sqrt(2)`, therefore `sin x =1//sqrt(2)` ...(ii)
`rArr x=3 pi//4` and so the general value of x is given by
`x=2n pi + 3pi//4, n in Z`
d. `log_(5) tan x=(log_(5) 4) (log_(4) (3 sin x))`
`rArr log_(5) tan x=log_(5) (3 sin x)` ...(i)
Since `log x` is real when `x gt 0`, we have
`tan x gt 0, sin x gt 0`
Therefore, x lies in the first quadrant. Now Eq. (i) gives
`tan x=3 sin x or cos x =1//3`
`:. x=2npi+ cos^(-1) (1//3), n in Z`