Question

Which of the below gives the value of $\int\limits_ {\dfrac {\pi} {{12}}} ^ {\dfrac {\pi}{4}} {\dfrac {{8\cos 2x}} {{{{(\tan x + \cot x)} ^3}}}} dx$:
A $\dfrac {{15}} {{128}}$
B $\dfrac {{15}} {{64}}$
C $\dfrac {{13}} {{32}}$
D $\dfrac {{13}} {{256}}$

Answer 1

In this question we have been given a trigonometric equation $\int\limits_ {\dfrac {\pi} {{12}}} ^ {\dfrac {\pi}{4}} {\dfrac {{8\cos 2x}} {{{{(\tan x + \cot x)} ^3}}}} dx$ we need to find its value. For that firstly we will change the tan and cot function into cosine and sin, after that we will be using the identity ${\sin ^2}x + {\cos ^2}x = 1$ for simplifying the equation. After that we will be using the formula: $2{\cos ^2}x - 1 = \cos 2x$ and then integrate using the substitution method, put the limits and solve the equation accordingly.

Complete step by step answer:
We want to find integration of a trigonometric expression. Consider $\int\limits_{\dfrac{\pi }{12}}^{\dfrac{\pi }{4}}{\dfrac{8\cos 2x}{{{(\tan x+\cot x)}^{3}}}}dx$,
For solving this equation in the first step we will change cot function and tan functions, as we know $\tan x=\dfrac{\sin x}{\cos x}$ and $\cot x=\dfrac{\cos x}{\sin x}$.
Now the equation would become $\int\limits_{\dfrac{\pi }{12}}^{\dfrac{\pi }{4}}{\dfrac{8\cos 2x}{{{\left( \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x} \right)}^{3}}}}dx$ ,
Now we will be taking the LCM and the equation would be $\int\limits_{\dfrac{\pi }{12}}^{\dfrac{\pi }{4}}{\dfrac{8\cos 2x}{{{\left( \dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x\cos x} \right)}^{3}}}}dx$,

Now we will be using the identity: ${{\sin }^{2}}x+{{\cos }^{2}}x=1$,
So, the equation would become, $\int\limits_{\dfrac{\pi }{12}}^{\dfrac{\pi }{4}}{\dfrac{8\cos 2x}{{{\left( \dfrac{1}{\sin x\cos x} \right)}^{3}}}}dx$,
Now we will take sin x and cos x in numerator to solve it further,
The equation would become, $\int\limits_{\dfrac{\pi }{12}}^{\dfrac{\pi }{4}}{8\cos 2x{{\sin }^{3}}x{{\cos }^{3}}x}dx$,
We can write this expression as $\int\limits_{\dfrac{\pi }{12}}^{\dfrac{\pi }{4}}{\cos 2x{{\left( 2\sin x\cos x \right)}^{3}}}dx$
Now we know that $\sin 2x=2\sin x\cos x$ .
Hence we get the expression as $\int\limits_{\dfrac{\pi }{12}}^{\dfrac{\pi }{4}}{\cos 2x{{\left( \sin 2x \right)}^{3}}}dx$
Now let us substitute $\sin 2x=t$ Differentiating on both sides we get $2\cos 2xdx=dt$
Hence we have when $\sin 2x=t$ $\cos 2xdx=\dfrac{dt}{2}$
Also when $x=\dfrac{\pi }{4},2x=\dfrac{\pi }{2}$ and hence $t=\sin 2x=1$
And when $x=\dfrac{\pi }{12},2x=\dfrac{\pi }{6}$ and hence $t=\sin 2x=\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
Now substituting the values and changing the limits we get
$\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{1}{2}{{\left( t \right)}^{3}}}dt$
Now we know that $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C$ hence we get.

& \int\limits_{\dfrac{1}{2}}^{1}{\dfrac{1}{2}{{\left( t \right)}^{3}}}dt=\dfrac{1}{2}\left[ \dfrac{{{t}^{4}}}{4} \right]_{\dfrac{1}{2}}^{1} \\\ & \int\limits_{\dfrac{1}{2}}^{1}{\dfrac{1}{2}{{\left( t \right)}^{3}}}dt=\dfrac{1}{2}\left[ \dfrac{{{1}^{4}}}{4}-{{\left( \dfrac{1}{2} \right)}^{4}}\times \dfrac{1}{4} \right] \\\ & \int\limits_{\dfrac{1}{2}}^{1}{\dfrac{1}{2}{{\left( t \right)}^{3}}}dt=\dfrac{1}{2}\left[ \dfrac{1}{4}-\dfrac{1}{64} \right] \\\ \end{aligned}$$ Now taking LCM we get $$\begin{aligned} & =\dfrac{1}{2}\left[ \dfrac{16-1}{64} \right] \\\ & =\dfrac{1}{2}\left[ \dfrac{15}{64} \right] \\\ & =\dfrac{15}{128} \\\ \end{aligned}$$ Hence we have The value of given Integral is $\dfrac{15}{128}$ **So, the correct answer is “Option A”.** **Note:** In this question a lot of formulas have been used, so keep all the formulas accordingly. While putting limits be careful as chances of mistakes are a lot over there. In the first step, students might expand ${\left( {\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}} \right)^3}$using the formula ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab(a + b)$, but avoid doing that as it will increase the calculations and chances of mistakes too. Also, if you are integrating by substitution method do change the limits accordingly.