Integral Solution: \int \frac{\sin x}{\sqrt{1+\cos x}} dx$
To find the integral $\int \frac{\sin x}{\sqrt{1+\cos x}} dx$, we can use trigonometric identities and simplify the integrand.
Simplify Integrand Using Trigonometric Identities
- Numerator: Use the double-angle identity for sine: $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$.
- Denominator: Use the half-angle identity for cosine: $1 + \cos x = 2 \cos^2 \frac{x}{2}$. Therefore, $\sqrt{1+\cos x} = \sqrt{2 \cos^2 \frac{x}{2}} = \sqrt{2} |\cos \frac{x}{2}|$.
Substitute these identities into the integral:
$ \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sqrt{2} |\cos \frac{x}{2}|} dx $Assuming $\cos \frac{x}{2} \ge 0$ (a common simplification in calculus problems), we have $|\cos \frac{x}{2}| = \cos \frac{x}{2}$. The integral simplifies to:
$ \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sqrt{2} \cos \frac{x}{2}} dx = \int \frac{2}{\sqrt{2}} \sin \frac{x}{2} dx = \int \sqrt{2} \sin \frac{x}{2} dx $Perform Integration
Now, integrate the simplified expression:
$ \sqrt{2} \int \sin \frac{x}{2} dx $Using the rule $\int \sin(ax) dx = -\frac{1}{a} \cos(ax) + c$, where $a = 1/2$:
$ \sqrt{2} \left( \frac{-\cos \frac{x}{2}}{1/2} \right) + c $Simplify the expression:
$ \sqrt{2} \left( -2 \cos \frac{x}{2} \right) + c = -2\sqrt{2} \cos \frac{x}{2} + c $Final Answer
The integral evaluates to $-2\sqrt{2} \cos \frac{x}{2} + c$.
Important Questions from Indefinite Integrals
- What is \(U(x)\,V(x)\) equal to?
$X^4 + C$ is the general integral of
What is \(\rm \int \frac{dx}{x(x^2 + 1)}\) equal to?
If \(\rm \int \sqrt{1 - sin 2x} \space dx\) = A sinx + B cosx + C, where 0 < x < \(\frac{\pi}{4}\) , then which one of the following is correct?
What is \(\smallint {{\rm{e}}^{\sin {\rm{x}}}}\left( {\frac{{{\rm{x}}{{\cos }^3}{\rm{x}} - \sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}}} \right){\rm{dx}}\) equal to?
