Question

The correct integral of \(\displaystyle\int \frac{\sin x}{\sqrt{1+\cos x}}\,dx\) (assuming integral constant as \(c\)) is ______.

• A. \(-2\sqrt{2}\,\sin\dfrac{x}{2} + c\)
• B. \(-2\sqrt{2}\,\cos\dfrac{x}{2} + c\)
• C. \(-\dfrac{1}{\sqrt{2}}\cos\dfrac{x}{2} + c\)
• D. \(-\dfrac{1}{\sqrt{2}}\sin\dfrac{x}{2} + c\)

Hint:

Substitute u = 1 + cos x so du = -sin x dx, integrate the resulting power of u, then convert back using the half-angle identity 1+cos x = 2cos^2(x/2).

Answer 1

The Correct Option is B

An alternative path skips the u-substitution and converts everything into half-angle form right at the start. Using the identity $1+\cos x = 2\cos^2\left(\frac{x}{2}\right)$ and $\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$, the integrand becomes

\[ \frac{\sin x}{\sqrt{1+\cos x}} = \frac{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{\sqrt{2}\left|\cos\left(\frac{x}{2}\right)\right|} \]

Taking the cosine term as positive over the region of interest, the $\cos(x/2)$ in the numerator cancels with the one in the denominator, leaving

\[ \frac{2\sin\left(\frac{x}{2}\right)}{\sqrt{2}} = \sqrt{2}\,\sin\left(\frac{x}{2}\right) \]

Now the integral is just $\int \sqrt{2}\,\sin\left(\frac{x}{2}\right)dx$. Using $\int \sin(kx)\,dx = -\frac{1}{k}\cos(kx) + c$ with $k = \frac{1}{2}$, this integrates to $\sqrt{2}\cdot\left(-2\cos\left(\frac{x}{2}\right)\right) + c = -2\sqrt{2}\,\cos\left(\frac{x}{2}\right) + c$, matching the earlier result exactly.

\[\boxed{-2\sqrt{2}\,\cos\frac{x}{2} + c}\]