$$\int \frac{\mathrm dx}{\sqrt {\sin x} \sqrt{\cos^7 x}}$$
That's the given value. I was dividing denominator and numerator by $\cos^4 (x)$.
$$\int \frac{sec^4 \mathrm dx}{\sqrt{\frac{\sin x}{\cos x}} \cdot \sqrt{\frac{\cos^7x}{\cos x}}}$$ $$\int \frac{sec^4 \mathrm dx}{\sqrt{\tan x}\cdot \sqrt{\cos^6x}}$$
But, my book wrote that
$$\int\frac{\sec^4x \mathrm dx}{\sqrt{\tan x}}$$
I have read a note in my book's context. They wrote that
Note : If m+n is real number than, you have to divide by $\cos^{m+n}x$ in this type of integral $$\int \frac{\mathrm dx}{\sin^mx \cos^nx}$$ then, you have take $\tan x=z$.
Why? Can't I prove that?
Picking up from where you left off
$$\int \frac{(1 + \tan^2(x))\sec^2(x)}{\sqrt {\tan(x)}}dx$$
$u = \tan (x) \rightarrow \frac{du}{dx} = \sec^2(x) \rightarrow dx = \frac{1}{\sec^2(x)} du$
$$\int \frac{u^2 + 1}{\sqrt{u}} du$$
$$\int \left( u^{\frac{3}{2}} + \frac{1}{\sqrt{u}} \right)du$$
$$\int u^{\frac{3}{2}} du + \int \frac{1}{\sqrt{u}} du$$
I'll let you take it from here