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# If $\int {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$, where $k$ is constant of integration, then find the value of$A + B + C$.(1) $\dfrac{{21}}{5}$ (2) $\dfrac{{21}}{{10}}$ (3) $\dfrac{{16}}{5}$ (4) $\dfrac{7}{{10}}$

Hint:Using the trigonometric identity of double angles, transform the integral in form of secant and tangent function. Now use the identity ${\sec ^2}x = {\tan ^2}x + 1$to transform the secant squared function into a tangent function. Take a proper substitution as $\tan x = m \Rightarrow {\sec ^2}xdx = dm$ and further solve the integral to obtain an expression as required in the question.

Here we have to integrate the given integral and compare our result with ${\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$ to find the value of A, B and C
Let’s get started with the solution of the given integral. We can start with simplifying the radical using the double-angle formula, i.e. $\sin 2x = 2\sin x\cos x$
$\Rightarrow \int {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}} = \int {\dfrac{{dx}}{{{{\cos }^3}x\sqrt {2 \times 2\sin x\cos x} }}} = \int {\dfrac{{dx}}{{2{{\cos }^3}x\sqrt {\sin x\cos x} }}}$

Now in the denominator, multiply and divide with $\cos x$, which will give:
$\Rightarrow \int {\dfrac{{dx}}{{2{{\cos }^3}x\sqrt {\sin x\cos x} }}} = \int {\dfrac{{dx}}{{\dfrac{{\cos x \times 2{{\cos }^3}x\sqrt {\sin x\cos x} }}{{\cos x}}}}}$
We can take this $\cos x$ inside the radical and then transform it further as:
$\Rightarrow \int {\dfrac{{dx}}{{\dfrac{{\cos x \times 2{{\cos }^3}x\sqrt {\sin x\cos x} }}{{\cos x}}}}} = \int {\dfrac{{dx}}{{2{{\cos }^4}x\sqrt {\dfrac{{\sin x\cos x}}{{{{\cos }^2}x}}} }}} = \int {\dfrac{{dx}}{{2{{\cos }^4}x\sqrt {\dfrac{{\sin x}}{{\cos x}}} }}}$
Since we know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$ and the reciprocal of $\cos x$ is $\sec x$ which can be used in the above integral transforming it further: $\dfrac{1}{{\cos x}} = \sec x \Rightarrow \dfrac{1}{{{{\cos }^4}x}} = {\sec ^2}x \times {\sec ^2}x$
$\Rightarrow \int {\dfrac{{dx}}{{2{{\cos }^4}x\sqrt {\dfrac{{\sin x}}{{\cos x}}} }}} = \int {\dfrac{{{{\sec }^2}x \times {{\sec }^2}xdx}}{{2\sqrt {\tan x} }}}$
Now, let’s use the trigonometric identity ${\sec ^2}x = {\tan ^2}x + 1$ in the numerator and utilize the method of substitution, i.e. assume that $\tan x = m \Rightarrow d\left( {\tan x} \right) = dm \Rightarrow {\sec ^2}xdx = dm$
Therefore, we get the integral as:$\int {\dfrac{{{{\sec }^2}x \times {{\sec }^2}xdx}}{{2\sqrt {\tan x} }}} = \int {\dfrac{{\left( {{{\tan }^2}x + 1} \right){{\sec }^2}xdx}}{{2\sqrt {\tan x} }}} = \int {\dfrac{{\left( {{m^2} + 1} \right)dm}}{{2\sqrt m }}}$
So, now we get an integral with the variable as $'m'$and it can be further split into two parts:
$\Rightarrow \int {\dfrac{{\left( {{m^2} + 1} \right)dm}}{{2\sqrt m }}} = \int {\dfrac{{{m^2}}}{{2\sqrt m }}dm + \int {\dfrac{1}{{2\sqrt m }}dm = } } \dfrac{1}{2}\int {{m^{2 - \dfrac{1}{2}}}dm + \dfrac{1}{2}\int {{m^{\dfrac{{ - 1}}{2}}}dm} }$
Now, using some basic integrals $\int {{a^b}da = \dfrac{{{a^{b + 1}}}}{{b + 1}}} + C$ in the above two integrals, we can further evaluate them as:
$\Rightarrow \dfrac{1}{2}\int {{m^{2 - \dfrac{1}{2}}}dm + \dfrac{1}{2}\int {{m^{\dfrac{{ - 1}}{2}}}dm} } = \dfrac{1}{2}\int {{m^{\dfrac{3}{2}}}dm + \dfrac{1}{2}\int {{m^{\dfrac{{ - 1}}{2}}}dm} } = \dfrac{1}{2}\dfrac{{{m^{\dfrac{3}{2} + 1}}}}{{\left( {\dfrac{3}{2} + 1} \right)}} + \dfrac{1}{2}\dfrac{{{m^{\dfrac{{ - 1}}{2} + 1}}}}{{\left( {\dfrac{{ - 1}}{2} + 1} \right)}} + C$
$\Rightarrow \dfrac{1}{2}\dfrac{{{m^{\dfrac{3}{2} + 1}}}}{{\left( {\dfrac{3}{2} + 1} \right)}} + \dfrac{1}{2}\dfrac{{{m^{\dfrac{{ - 1}}{2} + 1}}}}{{\left( {\dfrac{{ - 1}}{2} + 1} \right)}} + C = \dfrac{1}{2} \times \dfrac{2}{5}{m^{\dfrac{5}{2}}} + \dfrac{1}{2} \times 2{m^{\dfrac{1}{2}}} + C = \dfrac{1}{5}{m^{\dfrac{5}{2}}} + {m^{\dfrac{1}{2}}} + C$
Hence, we got the integral solved in a variable $m$; let’s substitute the assumed value $\tan x = m$ in the result.
$\Rightarrow \dfrac{1}{5}{m^{\dfrac{5}{2}}} + {m^{\dfrac{1}{2}}} + C = \dfrac{1}{5}{\left( {\tan x} \right)^{\dfrac{5}{2}}} + {\left( {\tan x} \right)^{\dfrac{1}{2}}} + k…………………...(1)$
But we need to compare the result of the integration with ${\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$
$\Rightarrow$ If ${\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k$ is equal to $\dfrac{1}{5}{\left( {\tan x} \right)^{\dfrac{5}{2}}} + {\left( {\tan x} \right)^{\dfrac{1}{2}}} + k$, then the coefficient of the terms must also be the same in each of the expressions.
Therefore, we can see that $A = \dfrac{1}{2},B = \dfrac{5}{2}$ and $C = \dfrac{1}{5}$
Hence, the required value for the expression $A + B + C = \dfrac{1}{2} + \dfrac{5}{2} + \dfrac{1}{5} = \dfrac{6}{2} + \dfrac{1}{5} = 3 + \dfrac{1}{5} = \dfrac{{16}}{5}$

So, the correct answer is “Option C”.

Note:Always solve the integrals step by step. Make the use of trigonometric identities wisely. Be careful while using the substitution method in the integrals, as it requires you to differentiate both sides of the assumed value and then change the variable in the whole integral. The substitution was an important part of this solution.