Question

Evaluate the Integral\(\int\limits_0^{\frac{\pi }{6}} {\sqrt {1 + cos2x} } \,dx\)

Short answer

\(\int\limits_0^{\frac{\pi }{6}} {\sqrt {1 + cos2x} } \,dx = \frac{1}{{\sqrt 2 }}\)\(\)

When we write\(cos2x = 2co{s^2}x - 1\), the integral gets simplified and easier to evaluate.

Step-by-step solution

Next we can write \(\sqrt {2co{s^2}x}  = \sqrt 2 cosx\) which it even more easier to evaluate.

\(\begin{aligned}{l}I &= \int\limits_0^{\frac{\pi }{6}} {\sqrt {1 + co{s^2}x} } dx\\cos2x &= 2co{s^2}x - 1\\I &= \int\limits_0^{\frac{\pi }{6}} {\sqrt {1 + 2co{s^2}x - 1} } \\I &= \int\limits_0^{\frac{\pi }{6}} {\sqrt {2co{s^2}x} } \\I &= \int\limits_0^{\frac{\pi }{6}} {\sqrt 2 } cosx\end{aligned}\)

Step 2:Integrating The Equation

\(I = \sqrt 2 \int\limits_0^{\frac{\pi }{6}} {cosx} \)

We know, \(\int {cosx = sinx} \)

Also, \(0 \le x \le \frac{\pi }{6}\)

\(cos \ge 0\)

Therefore \(\left| {cosx} \right| = cos\left( x \right)\)

\(I = \sqrt 2 \left( {\left( {sin\left( x \right)} \right)_0^{\frac{\pi }{6}}} \right)\)

Evaluate the integral. We need to substitute the upper and lower limits in\(sin\left( x \right)\).

Hence, \(I = \int\limits_0^{\frac{\pi }{6}} {\sqrt {1 + cos2x} } dx = \frac{1}{{\sqrt 2 }}\)