CUET UG 2025 — Mathematics Arithmetic

For $x \in \left(0, \frac{\pi}{2}\right)$x∈(0,2π​), find $\displaystyle \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx$∫sin2x​sinx+cosx​dx

Using the double angle formula $\sin 2x = 2\sin x \cos x$sin2x=2sinxcosx:

$$\int \frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} dx$$∫2sinxcosx​sinx+cosx​dx

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Let $t = \sin x - \cos x$t=sinx−cosx

Differentiating:

$$\frac{dt}{dx} = \cos x + \sin x$$dxdt​=cosx+sinx

$$dt = (\sin x + \cos x)dx$$dt=(sinx+cosx)dx

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To express the denominator in terms of $t$t:

$$t^2 = (\sin x - \cos x)^2$$t2=(sinx−cosx)2

$$t^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x$$t2=sin2x−2sinxcosx+cos2x

$$t^2 = 1 - 2\sin x \cos x$$t2=1−2sinxcosx

Therefore:

$$2\sin x \cos x = 1 - t^2$$2sinxcosx=1−t2

$$\sqrt{2\sin x \cos x} = \sqrt{1 - t^2}$$2sinxcosx​=1−t2​

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The integral becomes:

$$\int \frac{dt}{\sqrt{1-t^2}}$$∫1−t2​dt​

$$= \sin^{-1}(t) + C$$=sin−1(t)+C

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Substituting back $t = \sin x - \cos x$t=sinx−cosx:

$$\sin^{-1}(\sin x - \cos x) + C$$sin−1(sinx−cosx)+C