Find the Definite Integral of a Trigonometric Function
Calculus math problem
Find the definite integral of √(1 + cos x) from 0 to π
Solution to the definite integral of the trigonometric function
1 + cos x = cos 0 + cos x
⇒ 1 + cos x = 2 cos ((x + 0)/2) cos ((0 – x)/2)
so, 1 + cos x = 2 cos (x/2) cos (-x/2)
⇒ 1 + cos(x/2) = 2cos² (-x/2)
so
\begin{aligned} \sqrt{1+ \cos x} &= \sqrt{2 \cos^2 \bigg(\frac{x}{2}\bigg)}\\ \\ \Rightarrow \sqrt{1+ \cos x} &= \sqrt{2}\cos \bigg(\frac{x}{2}\bigg)\\ \end{aligned}
Now we get
\begin{aligned} \\ \int{\sqrt{1+ \cos x} \ \ dx} &= \int \sqrt{2}\cos \bigg(\frac{x}{2}\bigg) \ dx \\ \\ &= \sqrt{2}\int\cos \bigg(\frac{x}{2}\bigg) \ dx \\ \\ &=\sqrt{2}\sin\bigg(\frac{x}{2}\bigg) \times2 +C\\ \\ \Rightarrow \int{\sqrt{1+ \cos x} \ \ dx} &=2\sqrt{2}\sin\bigg(\frac{x}{2}\bigg)+C\\ \\ \end{aligned}
Now we can find the definite Integral
\begin{aligned} \\ \int_{0}^{\pi}{\sqrt{1+ \cos x} \ \ dx} &= \bigg[2\sqrt{2}\sin\bigg(\frac{x}{2}\bigg) \bigg]_{0}^{\pi} \\ \\ &=2\sqrt{2}\sin\bigg(\frac{\pi}{2}\bigg)-2\sqrt{2}\sin\bigg(\frac{0}{2}\bigg) \\ \\ \Rightarrow \int_{0}^{\pi}{\sqrt{1+ \cos x} \ \ dx} &=2\sqrt{2}\\ \\ \end{aligned}
Now we can find the value of the definite integral is 2√2