Question

Select correct option from the given options

Answer 1

HELLO DEAR,

let I = $\sf{\int\limits^{\pi\over2}_0}$ 1/(1 + cotx).dx

I = $\sf{\int\limits^{\pi\over2}_0}$ [ 1/{1 + (cosx/sinx)}].dx

I = $\sf{\int\limits^{\pi\over2}_0}$ {sinx/(sinx + cosx)}.dx

I = 1/2$\sf{\int\limits^{\pi\over2}_0}$ [{(sinx + cosx) + (sinx - cosx)}/(sinx + cosx)].dx

I = 1/2$\sf{\int\limits^{\pi\over2}_0}$ [1.dx + (sinx - cosx)/(sinx + cosx).dx]

put (sinx + cosx) = t

=> -(sinx - cosx).dx = dt

Also, [if , x = π/2 => t = (1 + 0) = 1] AND, [if x = 0 => t = (0 + 1) = 1]

therefore,

I = $\sf{1/2[x]^{\pi/2}_0}$ - 1/2$\sf{\int\limits^1_1}$ dt/t

I = 1/2(π/2 - 0) - 1/2$\sf{[log|t|]^1_1}$

I = π/4 - 1/2log|1| + 1/2log|1|

I = π/4

HENCE, OPTION ( A ) IS CORRECT

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Dear Student:

$\frac{1}{1+cotx} =\frac{sinx}{sinx+cosx}=$

For next step onwards see the attachment:

We got option a is correct.