Evaluate \displaystyle \int_0^\frac{\pi}{6} \sqrt{1+\sec^4x} dx.

Question:

Evaluate {eq}\displaystyle \int_0^{\frac{\pi}{6}} \sqrt{1+\sec^4x} dx {/eq}.

Integration using Taylor Series Expansion:

Let us consider any function{eq}y=f(x) {/eq}, whose Taylor series expansion about any point {eq}x=a {/eq} is given by:

{eq}f(x) = f(a) + \frac{{f'(a)}}{{1!}}(x - a) + \frac{{f''(a)}}{{2!}}{(x - a)^2} + \frac{{f'''(a)}}{{3!}}{(x - a)^3} + ...... {/eq}

Again:

{eq}{T_n}(x) = f(a) + \frac{{f'(a)}}{{1!}}(x - a) + \frac{{f''(a)}}{{2!}}{(x - a)^2} + \frac{{f'''(a)}}{{3!}}{(x - a)^3} + ...... + \frac{{{f^{(n)}}(a)}}{{n!}}{(x - a)^n} {/eq}

Hence we can use the same concept to evaluate any integral up to a certain degree.

Answer and Explanation:

Here, in this case, the given integral can be solved as:

{eq}\eqalign{ I& = \int\limits_0^{\frac{\pi }{6}} {\sqrt {1 + {{\sec }^4}x} } dx \cr & = \int\limits_0^{\frac{\pi }{6}} {\sqrt {1 + \frac{1}{{{{\cos }^4}x}}} } dx \cr & = \int\limits_0^{\frac{\pi }{6}} {\sqrt {\frac{{1 + {{\cos }^4}x}}{{{{\cos }^4}x}}} } dx \cr & = \int\limits_0^{\frac{\pi }{6}} {{{\sec }^2}x\sqrt {1 + {{\cos }^4}x} } dx \cr & = \int\limits_0^{\frac{\pi }{6}} {{{\sec }^2}x{{\left( {1 + {{\cos }^4}x} \right)}^{\frac{1}{2}}}} dx \cr & = \int\limits_0^{\frac{\pi }{6}} {{{\sec }^2}x\left( {1 + \frac{1}{2}{{\cos }^4}x} \right)} dx\,\,\,\,\,\,\,\left[ {{\text{Using 1st order Taylor Series expansion}}} \right] \cr & = \int\limits_0^{\frac{\pi }{6}} {\left( {{{\sec }^2}x + \frac{1}{2}{{\cos }^2}x} \right)} dx \cr & = \int\limits_0^{\frac{\pi }{6}} {\left( {{{\sec }^2}x + \frac{1}{4}\left( {2{{\cos }^2}x} \right)} \right)} dx \cr & = \int\limits_0^{\frac{\pi }{6}} {\left( {{{\sec }^2}x + \frac{1}{4}\left( {1 + \cos 2x} \right)} \right)} dx \cr & = \int\limits_0^{\frac{\pi }{6}} {\left( {{{\sec }^2}x + \frac{1}{4}\left( {1 + \cos 2x} \right)} \right)} dx \cr & = \left[ {\tan \left( x \right)} \right]_0^{\frac{\pi }{6}} + \frac{1}{2} \cdot \frac{1}{2}\left( {\int_0^{\frac{\pi }{6}} 1 dx + \int_0^{\frac{\pi }{6}} {\cos } \left( {2x} \right)dx} \right) \cr & = \frac{1}{{\sqrt 3 }} + \frac{{2\pi + 3\sqrt 3 }}{{48}} \cr & = \frac{{2\sqrt 3 \pi + 57}}{{48\sqrt 3 }} \cr} {/eq}


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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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