Question

Find the value of the following $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} } \right)$

Answer 1

In this question we need to find the integration of $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} } \right)$, for that firstly we will use the formula: $\sin 2x = 2\sin x\cos x$. After that we will change the cosine into sec using reciprocal identity to simplify the denominator. Now we will be also use identities like ${\sec ^2}x = 1 + {\tan ^2}x$ and $\dfrac{{\sin x}}{{\cos x}} = \tan x$. Next, we will find the integration using the substitution method by taking tanx = t and then evaluate it further.

Complete step-by-step answer:
We have been asked to find the integration of $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} } \right)$,
So, firstly in order to find the integration we will change sin2x using the formula: $\sin 2x = 2\sin x\cos x$,
Now it becomes: $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2 \times 2\sin x\cos x} }}} } \right)$
Now we will multiply and divide by cos x in the denominator,
So, we will get: $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{{{\cos }^4}x\sqrt {\dfrac{{2 \times 2\sin x\cos x}}{{\cos x}}} }}} } \right)$
Now as we know $\sec x = \dfrac{1}{{\cos x}}$
So, the equation will become, $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{{\sec }^4}xdx}}{{\sqrt {\dfrac{{2 \times 2\sin x\cos x}}{{\cos x}}} }}} } \right)$
Now we will write ${\sec ^4}x = {\sec ^2}x.{\sec ^2}x$
So, the equation would become, $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{{\sec }^2}x.{{\sec }^2}xdx}}{{\sqrt {\dfrac{{2 \times 2\sin x\cos x}}{{\cos x}}} }}} } \right)$
Now we will be using the identity, ${\sec ^2}x = 1 + {\tan ^2}x$ and we know $\dfrac{{\sin x}}{{\cos x}} = \tan x$
So, the equation would be: $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{({{\sec }^2}x.(1 + {{\tan }^2}x))dx}}{{\sqrt {4\tan x} }}} } \right)$
Now let tan x=t,
Differentiate with respect to x,
So, ${\sec ^2}xdx = dt$,
Now keep these values in $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{({{\sec }^2}x.(1 + {{\tan }^2}x))dx}}{{\sqrt {4\tan x} }}} } \right)$,
The equation would be: $\left( {\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{(1 + {\operatorname{t} ^2})dt}}{{\sqrt {4t} }}} } \right)$
Now we will take 4 out of the root,
So, the equation would be: $\left( {\dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{(1 + {\operatorname{t} ^2})dt}}{{\sqrt t }}} } \right)$
Now we will divide the numerator by the denominator,
The equation would be: $\left( {\dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{4}} {({t^{\dfrac{{ - 1}}{2}}}} + {t^{2 - \dfrac{1}{2}}})dt} \right)$
Now simplifying the equation: $\left( {\dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{4}} {({t^{\dfrac{{ - 1}}{2}}}} + {t^{\dfrac{3}{2}}})dt} \right)$
Now integrating the equation using the formula:$\int {{x^n}dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$
The limits would also change as if we put the limits in tan x=t. when we will keep the upper limit that is $\dfrac{\pi }{4}$in place of x we will get 1 and when the lower limit which is 0 we will get 0.
The equation would be:$\dfrac{1}{2}{\left[ {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + \dfrac{{{t^{\dfrac{3}{2} + 1}}}}{{\dfrac{3}{2} + 1}}} \right]^1}_0$
Solving it further,$\dfrac{1}{2}{\left[ {\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + \dfrac{{t\dfrac{5}{2}}}{{\dfrac{5}{2}}}} \right]^1}_0$
So, the value after simplification comes out to be:$\dfrac{2}{2}{\left[ {{t^{\dfrac{1}{2}}} + \dfrac{{{t^{\dfrac{5}{2}}}}}{5}} \right]^1}_0$
Now we will put the limits,$\left[ {\left( {1 + \dfrac{1}{5}} \right) - \left( {0 + 0} \right)} \right]$
So, the answer would be-$\dfrac{6}{5}$

Note: In this question we can keep the value of t as tan x after integrating but then do not change the limits as per tan x = t. Also be careful while keeping the limits as the chances of mistakes are more over there. In the beginning do change sin2x to 2sinxcosx otherwise it will become lengthy.