\((\pi / 2) \int_{0} \sqrt{(\sec x+1) d x}\) is equal to \(\ldots \ldots\) (a) 0 (b) \((\pi / 4)\) (c) \((\pi / 2)\) (d) \(\pi\)

Short Answer

Expert verified
(c) \((\pi / 2)\)

Step by step solution

01

Perform the trigonometric substitution

Substitute the given trigonometric identity: \(\sec{x} = \frac{1}{\cos{x}}\) in the integrand, so we get: \((\pi / 2) \int_{0} \sqrt{(\frac{1}{\cos{x}} + 1) dx}\)
02

Simplify the integrand

Now, let's simplify the integrand more by combining fractions inside the square root: \((\pi / 2) \int_{0} \sqrt{(\frac{1+\cos{x}}{\cos{x}}) dx}\)
03

Use the Pythagorean identity

We can further simplify the integrand by noticing that the numerator can be written in terms of sine using the Pythagorean identity \(1 - \sin^2{x} = \cos^2{x}\). So, the integrand becomes: \((\pi / 2) \int_{0} \sqrt{(\frac{2\sin^2{x/2}}{\cos{x}}) dx}\)
04

Cancel terms in the integrand

Now divide both the numerator and denominator by \(\cos{x}\). For this, we need to remember the double angle identity \(\sin{x} = 2\sin(x/2)\cos(x/2)\). After cancelation we get: \((\pi / 2) \int_{0} \sqrt{(4\sin^2{x/2}\cos^2{x/2}) dx}\)
05

Perform integration

Upon integrating the simplified integral, the result is: \((\pi / 2) \int_{0} (2\sin{x/2}\cos{x/2}) dx\) Using substitution, let \(u = x/2\), so \(du = dx/2\), and the bounds of integration become \(0 /2 = 0\) and \(\pi / 4\): \(2(\pi / 2) \int_{0}^{\pi / 4} (\sin{u}\cos{u}) (2 du)\) Simplifying further, we have: \(2\pi \int_{0}^{\pi / 4} (\sin{u}\cos{u}) d u\)
06

Evaluate the definite integral

Now integrate and evaluate \(2\pi [(\frac{1}{2} \sin^2{u})|_0^{\pi / 4}\) \(2\pi (\frac{1}{2} \sin^2{(\pi/4)} - \frac{1}{2} \sin^2{(0)})\) Plugging the values for the sine functions, we get: \(2\pi (\frac{1}{2}(1/2) - 0)\), Which simplifies to: \((\pi / 2)\) Thus, the answer is (c) \((\pi / 2)\).

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