Question

How to solve this ?? and explain it please :(
∫$\sqrt{ 2+ 2COS X}$

Answer 1

Answer:

$\sqrt{2} (2) \sqrt{(1 - cosx)}$ + Constant

Step-by-step explanation:

Step 1: Remove the common factor from the radicand

$\sqrt {2 + 2 cosx} = \sqrt{2} * \sqrt {1 + cosx}$

move the constant in front of the integral

$\sqrt {2} \int\ {\sqrt {1 + cosx} } \, dx$

Step 2: multiply integral by 1 - cosx

$\sqrt{2} \int{ (1 + cosx}) dx * \frac{(1 - cosx)}{(1 - cosx)} = \sqrt{2} \int {\sqrt{ \frac{ 1 - cos^2x}{1 - cosx } x} \, dx$

1 - $cos^2x$ is equivalent to $sin^2x$ by trig identities

Step 3: simplify the integral with the identity before using u-substitution

$\sqrt{2} \int {\sqrt{ \frac{ sin^2x}{1 - cosx} x} \, dx$

Step 4: Use U-Substitution

u = 1 - cosx

du = sinx dx,   dx = $\frac{du}{sinx}$

$\sqrt{2} \int {\frac{sinx}{\sqrt{u}}} \, \frac{du}{sinx} = \sqrt{2} \int {\frac{du}{\sqrt{u}} }$

Step 5: Integrate

$\int{u ^ \frac{-1}{2}} \, du = 2 u^\frac{1}{2}$

Step 6: substitute original u value and solve for final answer

 $\sqrt {2} (2) \sqrt{1 - cosx}$ + Constant