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

What is the definite integral of sec4x{\sec ^4}x from 00 to π4\dfrac{\pi }{4} ?

Explanation

Solution

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

Complete step-by-step solution:
Here, given that the function f(x)=sec4xf\left( x \right) = {\sec ^4}x is continuous under the closed interval [0,π4]\left[ {0,\dfrac{\pi }{4}} \right].
First, we are setting up the integral notation.
That is, 0π4sec4(x)dx\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
0π4sec4(x)dx=0π4sec2(x)sec2(x)dx\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, sec2(x)=tan2(x)+1{\sec ^2}\left( x \right) = {\tan ^2}\left( x \right) + 1
Substitute this identity for one of the sec2(x){\sec ^2}\left( x \right) . That is,
0π4sec4(x)dx=0π4[tan2(x)+1]sec2(x)dx\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 dydxtan(x)=sec2(x)\dfrac{{dy}}{{dx}}\tan \left( x \right) = {\sec ^2}\left( x \right),
So, let u=tan(x)u = \tan \left( x \right)
After differentiation, we get du=sec2(x)dxdu = {\sec ^2}\left( x \right)dx
Substitute these two in the main equation.
That is, 0π4sec4(x)dx=0π4[u2+1]du\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx = \int_0^{\dfrac{\pi }{4}} {\left[ {{u^2} + 1} \right]du}
=[u33+u]0π4= \left[ {\dfrac{{{u^3}}}{3} + u} \right]_0^{\dfrac{\pi }{4}}
Convert it into the terms of xx . That is, substitute u=tan(x)u = \tan \left( x \right)
0π4sec4(x)dx=[tan(x)33+tan(x)]0π4\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 xx . Then, subtracting the lower limit from the upper limit,

=[tan(π4)33+tan(π4)(tan(0)33+tan(0))] =[(1)33+1(0+0)] =[13+1] =[13+33] =[43] =1.3333  = \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, 0π4sec4(x)dx=1.3333\int_0^{\dfrac{\pi }{4}} {{{\sec }^4}\left( x \right)} dx = 1.3333
So, the answer is 1.33331.3333

Note: Note that sec2(x)=tan2(x)+1{\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.