Q88 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics

$\int \sqrt{1+\sin x}\,dx$ is equal to

  • Correct answer
    A
    $2\sqrt{2}\sin\left(\dfrac{x}{2} - \dfrac{\pi}{4}\right) + c$
  • B
    $2\sqrt{2}\sin\left(\dfrac{x}{2} + \dfrac{\pi}{4}\right) + c$
  • C
    $\sqrt{2}\sin\left(\dfrac{x}{2} - \dfrac{\pi}{4}\right) + c$
  • D
    $\sqrt{2}\sin\left(\dfrac{x}{2} + \dfrac{\pi}{4}\right) + c$

Solution

$I = \int \sqrt{1+\sin x}\,dx$

Using the trigonometric identities $\sin^2\left(\dfrac{x}{2}\right) + \cos^2\left(\dfrac{x}{2}\right) = 1$ and $\sin x = 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$, we get:

$1 + \sin x = \sin^2\left(\dfrac{x}{2}\right) + \cos^2\left(\dfrac{x}{2}\right) + 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right) = \left(\sin\left(\dfrac{x}{2}\right) + \cos\left(\dfrac{x}{2}\right)\right)^2$

Substituting this into the integral:

$I = \int \left(\sin\left(\dfrac{x}{2}\right) + \cos\left(\dfrac{x}{2}\right)\right)\,dx$

Integrating term by term:

$I = -2\cos\left(\dfrac{x}{2}\right) + 2\sin\left(\dfrac{x}{2}\right) + c$

$I = 2\left(\sin\left(\dfrac{x}{2}\right) - \cos\left(\dfrac{x}{2}\right)\right) + c$

Multiplying and dividing by $\sqrt{2}$:

$I = 2\sqrt{2}\left(\sin\left(\dfrac{x}{2}\right)\dfrac{1}{\sqrt{2}} - \cos\left(\dfrac{x}{2}\right)\dfrac{1}{\sqrt{2}}\right) + c$

$I = 2\sqrt{2}\left(\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{\pi}{4}\right) - \cos\left(\dfrac{x}{2}\right)\sin\left(\dfrac{\pi}{4}\right)\right) + c$

Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:

$I = 2\sqrt{2}\sin\left(\dfrac{x}{2} - \dfrac{\pi}{4}\right) + c$

Answer: $2\sqrt{2}\sin\left(\dfrac{x}{2} - \dfrac{\pi}{4}\right) + c$

Hence, the correct answer is option: 'A'.

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