The following integral `int_(pi/4)^(pi/2)(2cos e cx)^(17)dx` is equal to `(a)int_0^("log"(1+sqrt(2)))2(e^u+e^(-u))^(16)d u` `(b)int_0^("log"(1+sqrt(2)))2(e^u+e^(-u))^(17)d u` `(c)int_0^("log"(1+sqrt(2)))2(e^u-e^(-u))^(17)d u` `(d)int_0^("log"(1+sqrt(2)))2(e^u-e^(-u))^(16)d u`

Options

A. `overset(log(1+sqrt(2)))underset(0)int 2(e^(u)+e^(-u))^(16)du`
B. `overset(log(1+sqrt(2)))underset(0)int (e^(u)+e^(-u))^(17)du`
C. `overset(log(1+sqrt(2)))underset(0)int (e^(u)-e^(-u))^(17)du`
D. `overset(log(1+sqrt(2)))underset(0)int 2(e^(u)-e^(-u))^(16)du`

Answer

A

Explanation

Let `I=overset(pi//2)underset(pi//4)int (2 cosec x)^(17)dx` and let `e^(u)+e^(-u)=2 cosec x` Then,
`x=(pi)/(4)rArr e^(u)+e^(-u)=2sqrt(2)`
`rArr(e^(u))^(2)-2sqrt(2)e^(u)+1=0`
`rArr e^(u)=sqrt(2)+1 rArr u=log(1+sqrt(2))x=(pi)/(2)rArr e^(u)+e^(-u)=2`
Now, `e^(u)+e^(-u)=2 cosecx`
`rArr(e^(u))^(2)-(2 cosec x)e^(u)+=0`
`e^(u)=cosecx +cot x`
`rArr e^(-u)=(1)/(cosec x+cot x)=cosec x-cot x`
`:.cotx=(e^(u)-e^(-u))/(2)`
Again,`e^(u)+e^(-u)=2 cosec x`
`rArr (e^(u)-e^(-u))du=-2 cosec x cot x dx`
`rArr dx=-(e^(u)-e^(-u))/(2 cosec x cot x)du`
`rArr dx=-((e^(u)-e^(-u)))/((e^(u)+e^(-u))((e^(u)-e^(-u))/(2)))du`
`rArr dx=(-2)/(e^(u)+e^(-u))du`
`:. I=overset(pi//2)underset(pi//4)int (2 cosec x)^(17)dx`
`rArr I=overset(0)underset(log(1+sqrt(2)))int (e^(u)+e^(-u))^(17)xx(-2)/((e^(u)+e^(-u)))du`
`rArr I=overset(log(1+sqrt(2)))underset(2)int 2(e^(u)+e^(-u))^(16)du`