Interesting Integrals 7

Relevant wiki: Integration of Trigonometric Functions - Intermediate

Call the integration $\mathfrak{I}$ and use $\small{\color{teal}{\sin 2x=\dfrac{2\tan x }{\underbrace{1+\tan^2 x}_{\color{#D61F06}{\sec^2 x}}}}}$

$\mathfrak I=\dfrac{1}{2}\displaystyle\int\dfrac{\sec^4 x~\mathrm{d}x}{\sqrt{\tan x}}=\dfrac{1}{2}\displaystyle\int\dfrac{\sec^2 x(1+\tan^2 x)~\mathrm{d}x}{\sqrt{\tan x}}$

Substitute $\tan x=t^2$ such that $\sec^2 x~\mathrm{d}x=2t\mathrm{d}t$ and $1+\tan^2 x=1+t^4$ .

$\implies \mathfrak{I}=\dfrac 12\displaystyle\int\dfrac{(1+t^4)2t~\mathrm{d}t}{t}$

$=\displaystyle\int (1+t^4)\mathrm{d}t$

$=t+\dfrac{t^5}5+k=\tan x^{1/2}+\dfrac 15\cdot\tan x^{5/2}+k$

Hence, $1/2+5/2+1/5=16/5$

So that $\Large 16+5=\boxed{21}$

Relevant wiki: Integration of Trigonometric Functions - Intermediate

$\begin{aligned} I & = \int \frac {dx}{\cos^3 x \sqrt{2\sin2x}} \\ & = \int \frac {dx}{\cos^3 x \sqrt{4\sin x \cos x}} \\ & = \int \frac {dx}{2\cos^4 x \sqrt{\frac {\sin x}{\cos x}}} \\ & = \int \frac {\sec^4 x}{2 \tan^\frac 12 x} dx & \small \color{#3D99F6}{\text{Let }t = \tan x \implies dt = \sec^2 \ dx \implies \sec^2 x = 1+t^2} \\ & = \int \frac {1+t^2}{2t^\frac 12} dt \\ & = \frac 12 \int t^{-\frac 12} \ dt + \frac 12 \int t^\frac 32 \ dt \\ & = t^\frac 12 + \frac 15 t^\frac 52 + k & \small \color{#3D99F6}{\text{Putting back }\tan x = t} \\ & = (\tan x)^\frac 12 + \frac 15 (\tan x)^\frac 52 + k\end{aligned}$

$\implies A + B + C = \frac 12 + \frac 52 + \frac 15 = \frac {16}5 \implies p+q = 16+5 = \boxed{21}$ .

$\displaystyle \int \dfrac{dx}{\cos^3 x\sqrt{2\sin2x}}\\ =\displaystyle \int \dfrac{dx}{\cos^3 x\sqrt{\color{#3D99F6}{4\sin x \cos x}}}\\ =\displaystyle \int \dfrac{dx}{2\cos^3 x\sqrt{\frac{\sin x \cos^2 x}{\cos x}}}\\ =\displaystyle \int \dfrac{dx}{2\cos^4 x\sqrt{\color{#D61F06}{\tan x}}}\\ =\displaystyle \int \dfrac{\color{#EC7300}{\sec^4 x}}{2\sqrt{\tan x}} dx$

Let $\color{#69047E}{u=\sqrt{\tan x}} \implies u^4=\color{#20A900}{\tan^2 x = \sec^2 x - 1} \implies \color{magenta}{\sec^2 x = u^4+1}$

$\dfrac{du}{dx}=\dfrac{1}{2\sqrt{\tan x}}\sec^2 x\\ \implies \color{teal}{du = \dfrac{\sec^2 x}{2\sqrt{\tan x}} dx}$

$\displaystyle \int \dfrac{\sec^4 x}{2\sqrt{\tan x}} dx\\ =\displaystyle \int \color{teal}{\dfrac{\color{magenta}{\sec^2 x}\sec^2 x}{2\sqrt{\tan x}} dx}\\ =\displaystyle \int \color{magenta}{u^4+1}\;\color{teal}{du}\\ =\dfrac{\color{#69047E}{u}^5}{5}+\color{#69047E}{u}+k\\ =\dfrac{1}{5}\left(\color{#69047E}{\sqrt{\tan x}}\right)^5+\color{#69047E}{\sqrt{\tan x}}+k\\ =\left(\tan x\right)^{1/2}+\dfrac{1}{5}\left(\tan x\right)^{5/2}+k$

$A=\dfrac{1}{2},\;B=\dfrac{5}{2},\;C=\dfrac{1}{5},\;A+B+C=\dfrac{1}{2}+\dfrac{5}{2}+\dfrac{1}{5}=\dfrac{16}{5}\\ \implies p=16,\;q=5,\;p+q=16+5=\boxed{21}$

Theorems used:

1. Double angle formula : $\color{#3D99F6}{\sin2x = 2\sin x\cos x}$
2. Definition of tangent : $\color{#D61F06}{\tan x= \dfrac{\sin x}{\cos x}}$
3. Definition of secant : $\color{#EC7300}{\dfrac{1}{\cos x} = \sec x}$
4. Pythagorean trigonometric identity : $\color{#20A900}{\tan^2 x +1 = \sec^2 x}$