JEE Challenger
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MathHardAdv 2008 • Paper 2

Integral Representation of Area Between Trigonometric Square Root Curves

The area of the region between the curves y=1+sin⁡xcos⁡xy = \sqrt{\frac{1 + \sin x}{\cos x}} and y=1−sin⁡xcos⁡xy = \sqrt{\frac{1 - \sin x}{\cos x}} bounded by the lines x=0x = 0 and x=π4x = \frac{\pi}{4} is

Options

A

∫02−1t(1+t2)1−t2 dt\int_0^{\sqrt{2}-1} \frac{t}{(1+t^2)\sqrt{1-t^2}}\,dt

B

∫02−14t(1+t2)1−t2 dt\int_0^{\sqrt{2}-1} \frac{4t}{(1+t^2)\sqrt{1-t^2}}\,dt

Correct
C

∫02+14t(1+t2)1−t2 dt\int_0^{\sqrt{2}+1} \frac{4t}{(1+t^2)\sqrt{1-t^2}}\,dt

D

∫02+1t(1+t2)1−t2 dt\int_0^{\sqrt{2}+1} \frac{t}{(1+t^2)\sqrt{1-t^2}}\,dt

Topics & Concepts

Application of IntegralsArea Under CurvesDefinite Integration

Step-by-Step Solution

To find the area of the region bounded by the given curves and lines, we first determine the relative positions of the two curves on the interval x∈[0,π4]x \in \left[0, \frac{\pi}{4}\right].

The given curves are:
y1=1+sin⁡xcos⁡xandy2=1−sin⁡xcos⁡xy_1 = \sqrt{\frac{1 + \sin x}{\cos x}} \quad \text{and} \quad y_2 = \sqrt{\frac{1 - \sin x}{\cos x}}

For x∈[0,π4]x \in \left[0, \frac{\pi}{4}\right], we have sin⁡x≥0\sin x \ge 0, which implies 1+sin⁡x≥1−sin⁡x1 + \sin x \ge 1 - \sin x. Since cos⁡x>0\cos x > 0 in this interval, it follows that y1≥y2y_1 \ge y_2.

Thus, the area AA of the bounded region is given by:
A=∫0π/4(y1−y2) dx=∫0π/4(1+sin⁡xcos⁡x−1−sin⁡xcos⁡x)dx=∫0π/41+sin⁡x−1−sin⁡xcos⁡x dxA = \int_0^{\pi/4} (y_1 - y_2) \, dx = \int_0^{\pi/4} \left( \sqrt{\frac{1 + \sin x}{\cos x}} - \sqrt{\frac{1 - \sin x}{\cos x}} \right) dx = \int_0^{\pi/4} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{\cos x}} \, dx

Next, we simplify the numerator using half-angle formulas:
1+sin⁡x=cos⁡2x2+sin⁡2x2+2sin⁡x2cos⁡x2=(cos⁡x2+sin⁡x2)21 + \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2
1−sin⁡x=cos⁡2x2+sin⁡2x2−2sin⁡x2cos⁡x2=(cos⁡x2−sin⁡x2)21 - \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2

For x∈[0,π4]x \in \left[0, \frac{\pi}{4}\right], we have x2∈[0,π8]\frac{x}{2} \in \left[0, \frac{\pi}{8}\right], so cos⁡x2>sin⁡x2≥0\cos\frac{x}{2} > \sin\frac{x}{2} \ge 0. Therefore:
1+sin⁡x=cos⁡x2+sin⁡x2\sqrt{1 + \sin x} = \cos\frac{x}{2} + \sin\frac{x}{2}
1−sin⁡x=cos⁡x2−sin⁡x2\sqrt{1 - \sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}

Subtracting the two expressions gives:
1+sin⁡x−1−sin⁡x=2sin⁡x2\sqrt{1 + \sin x} - \sqrt{1 - \sin x} = 2\sin\frac{x}{2}

So the integral becomes:
A=∫0π/42sin⁡(x/2)cos⁡x dxA = \int_0^{\pi/4} \frac{2\sin(x/2)}{\sqrt{\cos x}} \, dx

Now, perform the substitution t=tan⁡(x2)t = \tan\left(\frac{x}{2}\right):

  • When x=0x = 0, t=tan⁡(0)=0t = \tan(0) = 0.
  • When x=π4x = \frac{\pi}{4}, t=tan⁡(π8)=2−1t = \tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1.
  • The differential is dx=21+t2 dtdx = \frac{2}{1 + t^2} \, dt.
  • Trigonometric functions in terms of tt:
    sin⁡(x2)=t1+t2\sin\left(\frac{x}{2}\right) = \frac{t}{\sqrt{1 + t^2}}
    cos⁡x=1−t21+t2\cos x = \frac{1 - t^2}{1 + t^2}

Substitute these expressions into the integral:
A=∫02−12(t1+t2)1−t21+t2⋅21+t2 dtA = \int_0^{\sqrt{2}-1} \frac{2 \left(\frac{t}{\sqrt{1+t^2}}\right)}{\sqrt{\frac{1-t^2}{1+t^2}}} \cdot \frac{2}{1+t^2} \, dt

Simplifying the integrand:
A=∫02−12t1+t2⋅1+t21−t2⋅21+t2 dtA = \int_0^{\sqrt{2}-1} \frac{2t}{\sqrt{1+t^2}} \cdot \frac{\sqrt{1+t^2}}{\sqrt{1-t^2}} \cdot \frac{2}{1+t^2} \, dt
A=∫02−14t(1+t2)1−t2 dtA = \int_0^{\sqrt{2}-1} \frac{4t}{(1+t^2)\sqrt{1-t^2}} \, dt

This matches option (B).

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