Solution:
we have$\frac{\cos
x-\sin x}{\sqrt{\sin x\cos x}}=\frac{\cos x}{\sqrt{\sin x\cos x}}-\frac{\sin
x}{\sqrt{\sin x\cos x}}$ &$1+6\sin x\cos x=1+6\frac{\sin x}{\cos x}{{\cos
}^{2}}x$ $=1+6\frac{\tan x}{{{\sec }^{2}}x}$ so $\int{\frac{\frac{\cos x-\sin
x}{\sqrt{\sin x\cos x}}}{1+6\sin x\cos x}dx}=\int{\frac{\frac{\cos
x}{\sqrt{\sin x\cos x}}-\frac{\sin x}{\sqrt{\sin x\cos x}}}{1+6\frac{\tan
x}{{{\sec }^{2}}x}}dx}$
but $\frac{\cos
x}{\sqrt{\sin x\cos x}}=\frac{\cos x}{\sqrt{\frac{\cos x}{\sin x}{{\sin
}^{2}}x}}=\frac{\cos x}{\sqrt{\frac{\cot x}{{{\csc }^{2}}x}}}=\frac{\cos
x}{\frac{\sqrt{\cot x}}{\csc x}}=\frac{\csc x\cos x}{\sqrt{\cot x}}=\frac{\cot
x}{\sqrt{\cot x}}=\sqrt{\cot x}$ &$\frac{\sin
x}{\sqrt{\sin x\cos x}}=\frac{\sin x}{\sqrt{\frac{\sin x}{\cos x}{{\cos
}^{2}}x}}=\frac{\sin x}{\sqrt{\frac{\tan x}{{{\sec }^{2}}x}}}=\frac{\sin x\sec
x}{\sqrt{\tan x}}=\frac{\tan x}{\sqrt{\tan x}}=\sqrt{\tan x}$
hence $\int{\frac{\frac{\cos x-\sin x}{\sqrt{\sin x\cos x}}}{1+6\sin x\cos x}dx}$ \(=\int{\frac{\sqrt{\cot
x}-\sqrt{\tan x}}{1+6\frac{\tan x}{{{\sec }^{2}}x}}dx}=\int{\frac{{{\sec
}^{2}}x\left( \sqrt{\cot x}-\sqrt{\tan x} \right)}{{{\sec }^{2}}x+6\tan x}dx}\) but ${{\sec
}^{2}}x={{\tan }^{2}}x+1$ so $\int{\frac{{{\sec }^{2}}x\left( \sqrt{\cot
x}-\sqrt{\tan x} \right)}{{{\sec }^{2}}x+6\tan x}dx}=\int{\frac{\left( {{\tan
}^{2}}x+1 \right)\left( \sqrt{\cot x}-\sqrt{\tan x} \right)}{{{\tan
}^{2}}x+1+6\tan x}dx}$
\(=\int{\frac{\left(
{{\tan }^{2}}x+1 \right)\left( \sqrt{\cot x}-\sqrt{\tan x} \right)}{{{\tan
}^{2}}x+2\tan x+4\tan x+1}dx}=\int{\frac{{{\sec }^{2}}x\left( \sqrt{\cot
x}-\sqrt{\tan x} \right)}{{{\left( \tan x+1 \right)}^{2}}+4\tan x}dx}\) but $\sqrt{\cot
x}=\sqrt{\frac{1}{\tan x}}$ $\Rightarrow \sqrt{\cot x}-\sqrt{\tan
x}=\frac{1}{\sqrt{\tan x}}-\sqrt{\tan x}=\frac{1-\tan x}{\sqrt{\tan x}}$so $\int{\frac{{{\sec
}^{2}}x\left( \sqrt{\cot x}-\sqrt{\tan x} \right)}{{{\left( \tan x+1
\right)}^{2}}+4\tan x}dx}$ $=\int{\frac{{{\sec }^{2}}x\left( \frac{1-\tan
x}{\sqrt{\tan x}} \right)}{{{\left( \tan x+1 \right)}^{2}}+4\tan x}dx}$$=\int{\frac{\left(
1-\tan x \right)\frac{{{\sec }^{2}}x}{\sqrt{\tan x}}}{{{\left( \tan x+1
\right)}^{2}}+4\tan x}dx}$ $=-\int{\frac{\left( \tan x-1 \right)\frac{{{\sec
}^{2}}x}{\sqrt{\tan x}}}{{{\left( \tan x+1 \right)}^{2}}+4\tan x}dx}$ $=-\int{\frac{\left(
\tan x-1 \right)\frac{2{{\sec }^{2}}x}{2\sqrt{\tan x}}}{{{\left( \tan x+1
\right)}^{2}}+4\tan x}dx}=-2\int{\frac{\tan x-1\frac{{{\sec
}^{2}}x}{2\sqrt{\tan x}}}{{{\left( \tan x+1 \right)}^{2}}+4\tan x}dx}$
$=-2\int{\frac{\tan
x-1}{{{\left( \tan x+1 \right)}^{2}}+4\tan x}\frac{{{\sec }^{2}}x}{2\sqrt{\tan
x}}dx}$ Let $u=\sqrt{\tan x}\Rightarrow du=\frac{\left( \tan x
\right)'}{2\sqrt{\tan x}}=\frac{{{\sec }^{2}}x}{2\sqrt{\tan x}}dx$ & ${{u}^{2}}=\tan x$ so $-2\int{\frac{\tan
x-1}{{{\left( \tan x+1 \right)}^{2}}+4\tan x}\times \frac{{{\sec
}^{2}}x}{2\sqrt{\tan x}}dx}=-2\int{\frac{{{u}^{2}}-1}{{{\left( {{u}^{2}}+1
\right)}^{2}}+4{{u}^{2}}}du}$ $=-2\int{\frac{{{u}^{2}}\left(
1-\frac{1}{{{u}^{2}}} \right)}{{{\left( u\left( u+\frac{1}{u} \right)
\right)}^{2}}+4{{u}^{2}}}du}=-2\int{\frac{{{u}^{2}}\left( 1-\frac{1}{{{u}^{2}}}
\right)}{{{u}^{2}}\left[ {{\left( u+\frac{1}{u} \right)}^{2}}+4
\right]}du}=-2\int{\frac{1-\frac{1}{{{u}^{2}}}}{{{\left( u+\frac{1}{u}
\right)}^{2}}+4}du}$
Let $w=u+\frac{1}{u}\Rightarrow
dw=\left( 1-\frac{1}{{{u}^{2}}} \right)du$ so $-2\int{\frac{1-\frac{1}{{{u}^{2}}}}{{{\left(
u+\frac{1}{u} \right)}^{2}}+4}du}=-2\int{\frac{dw}{{{\left( 2
\right)}^{2}}+{{w}^{2}}}}=-2\int{\frac{dw}{4\left( 1+{{\left( \frac{w}{2}
\right)}^{2}} \right)}}$
Let $v=\frac{w}{2}\Rightarrow
2v=w\Rightarrow $ so $-2\int{\frac{dw}{4\left( 1+{{\left( \frac{w}{2}
\right)}^{2}} \right)}=-2\int{\frac{2dv}{4\left( 1+{{v}^{2}}
\right)}=-\int{\frac{dv}{1+{{v}^{2}}}}}}$ $=-\arctan \left( v
\right)+c=-\arctan \left( \frac{w}{2} \right)+c=-\arctan \left(
\frac{u+\frac{1}{u}}{2} \right)+c=-\arctan \left( \frac{{{u}^{2}}+1}{2u}
\right)+c$$=-\arctan \left( \frac{\tan x+1}{2\sqrt{\tan x}} \right)+c$
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