Question

What is the definite integral of \[{\sec ^4}x\] from \[0\] to \[\dfrac{\pi }{4}\] ?

Answer 1

Hint: In general, evaluating a definite integral entails determining the region enclosed by the function's graph and the \[x\] axis over the given interval \[\left[ {a,b} \right]\]. The area of the region in \[xy\] plane bounded by the graph of \[f\] , the \[x\] axis, and the lines \[x = a\] and \[x = b\] , so that area above the \[x\] axis adds to the total, and area below the \[x\] axis subtracts from the total, is shown using \[\int\limits_a^b {f\left( x \right)dx} \]

Complete step-by-step solution:
Here, given that the function \[f\left( x \right) = {\sec ^4}x\] is continuous under the closed interval \[\left[ {0,\dfrac{\pi }{4}} \right]\].
First, we are setting up the integral notation.
That is, \[\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx\]
Using the normal integration rules, we are going to find the integral. For that, we are splitting the function as
\[\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx = \int_0^{\dfrac{\pi }{4}} {{{\sec }^2}} \left( x \right){\sec ^2}\left( x \right)dx\]
We know that, \[{\sec ^2}\left( x \right) = {\tan ^2}\left( x \right) + 1\]
Substitute this identity for one of the \[{\sec ^2}\left( x \right)\] . That is,
\[\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx = \int_0^{\dfrac{\pi }{4}} {\left[ {{{\tan }^2}\left( x \right) + 1} \right]} {\sec ^2}\left( x \right)dx\]
Suppose we are solving an equation that includes both a function and its derivative. In that case, we will probably want to use u-substitution, which is also known as the integration by substitution method.
Here it is clearly visible that we have to follow this method since \[\dfrac{{dy}}{{dx}}\tan \left( x \right) = {\sec ^2}\left( x \right)\],
So, let \[u = \tan \left( x \right)\]
After differentiation, we get \[du = {\sec ^2}\left( x \right)dx\]
Substitute these two in the main equation.
That is, \[\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx = \int_0^{\dfrac{\pi }{4}} {\left[ {{u^2} + 1} \right]du} \]
\[ = \left[ {\dfrac{{{u^3}}}{3} + u} \right]_0^{\dfrac{\pi }{4}}\]
Convert it into the terms of \[x\] . That is, substitute \[u = \tan \left( x \right)\]
\[\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx = \left[ {\dfrac{{\tan {{\left( x \right)}^3}}}{3} + \tan \left( x \right)} \right]_0^{\dfrac{\pi }{4}}\]
Then continuing definite integral calculation, we are substituting the limit values to \[x\] . Then, subtracting the lower limit from the upper limit,
\[
   = \left[ {\dfrac{{\tan {{\left( {\dfrac{\pi }{4}} \right)}^3}}}{3} + \tan \left( {\dfrac{\pi }{4}} \right) - \left( {\dfrac{{\tan {{\left( 0 \right)}^3}}}{3} + \tan \left( 0 \right)} \right)} \right] \\
   = \left[ {\dfrac{{{{\left( 1 \right)}^3}}}{3} + 1 - \left( {0 + 0} \right)} \right] \\
   = \left[ {\dfrac{1}{3} + 1} \right] \\
   = \left[ {\dfrac{1}{3} + \dfrac{3}{3}} \right] \\
   = \left[ {\dfrac{4}{3}} \right] \\
   = 1.3333 \\
 \]
Therefore, \[\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx = 1.3333\]
So, the answer is \[1.3333\]

Note: Note that \[{\sec ^2}\left( x \right) = {\tan ^2}\left( x \right) + 1\] . Remember that when we are doing definite integration, we have to subtract the lower limit from the upper limit. Before solving the integration we have to simplify the expression as we can’t integrate the expression directly.