The area of the region between the curves y = √((1+sin x)/(cos x)) and y = √((1-sin x)/(cos x)) and bounded by the lines x = 0 and x = (π)/4 is

int_0^√(2)-1 dfrac4t(1+t^2)√(1-t^2) dt

int_0^√(2)-1 dfract(1+t^2)√(1-t^2) dt

int_0^√(2)+1 dfrac4t(1+t^2)√(1-t^2) dt

int_0^√(2)+1 dfract(1+t^2)√(1-t^2) dt

The area of the region between the curves y = √((1+sin… — JEE Main 2008 Mathematics PYQ

The area of the region between the curves

y = \sqrt{\dfrac{1+\sin x}{\cos x}}

and

y = \sqrt{\dfrac{1-\sin x}{\cos x}}

and bounded by the lines

x = 0

and

x = \dfrac{\pi}{4}

A.
$\int_0^{\sqrt{2}-1} \dfrac{t}{(1+t^2)\sqrt{1-t^2}}\, dt$
B.
$\int_0^{\sqrt{2}-1} \dfrac{4t}{(1+t^2)\sqrt{1-t^2}}\, dt$
C.
$\int_0^{\sqrt{2}+1} \dfrac{4t}{(1+t^2)\sqrt{1-t^2}}\, dt$
D.
$\int_0^{\sqrt{2}+1} \dfrac{t}{(1+t^2)\sqrt{1-t^2}}\, dt$

Show answer and solution

Correct option: B

Solution

The area is ∫₀^(π/4) [√((1+sinx)/cosx) - √((1-sinx)/cosx)] dx.

Using half-angle identities: (1+sinx)/cosx = (cos(x/2)+sin(x/2))²/cos²(x/2)-sin²(x/2)) simplifies so that √((1+sinx)/cosx) = √((1+tan(x/2))/(1-tan(x/2))) and similarly the difference becomes:

√((1+sinx)/cosx) - √((1-sinx)/cosx) = [cos(x/2)+sin(x/2) - cos(x/2)+sin(x/2)] / √(cos²(x/2)-sin²(x/2)) = 2sin(x/2)/√(cosx).

So the area = ∫₀^(π/4) 2sin(x/2)/√(cosx) dx.

Substitute t = tan(x/2), so x/2 = arctan t, dx = 2dt/(1+t²), sin(x/2) = t/√(1+t²), cosx = (1-t²)/(1+t²) so √(cosx) = √(1-t²)/√(1+t²).

When x=0, t=0; when x=π/4, t=tan(π/8) = √2-1.

The integrand becomes: 2·(t/√(1+t²)) / (√(1-t²)/√(1+t²)) · 2dt/(1+t²) = 4t/

[AIEEE 2008]

JEE Main 2008 Mathematics question on Definite Integrals and Areas