{1+ x} d x is equal to

Correct Answer is : 22sin⁡(x2−π4)+c2 {2} ≤ft(x/2-π/4)+c22sin(2x−4π)+c

Correct Answer is : 22sin⁡(x2−π4)+c2 {2} ≤ft(x/2-π/4)+c22sin(2x−4π)+c

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MHTCET 13 Apr 2026 Shift 2 Paper

Question : 130 of 150

Marks: +1, -0

\int \sqrt{1+\sin x} d x

is equal to

Solution:

Concept:

Use the identity

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

to simplify the integrand, then integrate and rewrite the result as a single sine function.

Explanation:

We know

\sin^2\frac{x}{2}+\cos^2\frac{x}{2}=1

and

2\sin\frac{x}{2}\cos\frac{x}{2}=\sin x

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

Taking the positive square root,

\sqrt{1+\sin x} = \sin\frac{x}{2}+\cos\frac{x}{2}

Integrate:

\int \left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)dx = -2\cos\frac{x}{2}+2\sin\frac{x}{2}+C

=2\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)+C

Use the identity

\sin A - \cos A = \sqrt{2}\sin\left(A-\frac{\pi}{4}\right)

with

A=\frac{x}{2}

Then

2\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)=2\sqrt{2}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)

Therefore

\int \sqrt{1+\sin x}\,dx = 2\sqrt{2}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)+C

Answer:

Option A:

2\sqrt{2}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)+c

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