Question

The value of the integral `int_(-1)^1log_e(sqrt(1 - x) + sqrt(1 + x))dx` is equal to ______.

पर्याय

• `2log_e2 + π/4 - 1`

• `1/2log_e2 + π/4 - 3/2`

• `2log_e2 + π/2 - 1/2`

• `log_e2 + π/2 - 1`

Answer 1

The value of the integral `int_(-1)^1log_e(sqrt(1 - x) + sqrt(1 + x))dx` is equal to `underlinebb(log_e2 + π/2 - 1)`.

Explanation:

Given f(x) = In `(sqrt(1 - x) + sqrt(1 + x))`

f(–x) = In `(sqrt(1 - (-x)) + sqrt(1 - x))`

⇒ f(–x) = In `(sqrt(1 + x) + sqrt(1 - x)) = f(x)`

⇒ f(–x) = f(x)

∴ f is even function

Now, I = `int_(-1)^1 In(sqrt(1 - x) + sqrt(1 + x))dx`

⇒ I = `2int_0^1 In(sqrt(1 - x) + sqrt(1 + x))dx`

Put, x = cos2θ ⇒ dx = –2sin2θdθ also cos2θ = 2cos²θ – 1 = 1 – 2sin²θ 

Limits, x = 0, θ = `π/4`

x = 1, θ = 0

⇒ I = `-4int_(π/4)^0 [In{(sinθ + cosθ)sqrt(2)}]sin2θdθ`

= `4int_0^(π/4) [In {(sinθ + cosθ)sqrt(2)}]\sin2θdθ`

= `4int_0^(π/4) In (sinθ + cosθ)sin2θdθ + 4  In sqrt(2) int_0^(π/4) sin2θdθ`

= `4[0 + 1/2 int_0^(π/4) (cosθ - sinθ)^2dθ] + 4  In sqrt(2)(0 + 1/2)`

= `4[0 + 1/2 int_0^(π/4) (1 - sin2θ)dθ] + 2 In sqrt(2)`

= `2[θ + (cos2θ)/2]_0^(π/4) + In 2`

= `2[π/4 - 1/2] + In 2`

∴ I = `π/2 - 1 + In 2`