Two integral exercises have the same idea

Exercise:

Integrate, $\int{{{\left( \sin x+\cos x \right)}^{9}}\cos 2x}\,dx$

Solution: we know that $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x=\left( \cos x+\sin x \right)\left( \cos x-\sin x \right)$

So $\int{{{\left( \sin x+\cos x \right)}^{9}}\cos 2x}\,dx=\int{{{\left( \sin x+\cos x \right)}^{9}}\left( \sin x+\cos x \right)\left( \cos x-\sin x \right)\,dx}$

$=\int{{{\left( \sin x+\cos x \right)}^{10}}\left( \cos x-\sin x \right)dx}$

Take $u=\sin x+\cos x\Leftrightarrow du=\left( \cos x-\sin x \right)dx$

So $\int{{{\left( \sin x+\cos x \right)}^{10}}\left( \cos x-\sin x \right)dx}=\int{{{\left( \cos x+\sin x \right)}^{10}}d\left( \cos x+\sin x \right)}$

$=\frac{1}{11}{{\left( \cos x+\sin x \right)}^{11}}+c$

Exercise:

Integrate, $\int{\frac{\csc x-\sec x}{1+\sec x\csc x}dx}$

Solution : we have $\csc x-\sec x=\frac{1}{\sin x}-\frac{1}{\cos x}=\frac{\cos x-\sin x}{\sin x\cos x}$

Also $1+\sec x\csc x=1+\frac{1}{\cos x\sin x}=\frac{\cos x\sin x+1}{\cos x\sin x}$

So $\frac{\csc x-\sec x}{1+\sec x\csc x}=\frac{\frac{\cos x-\sin x}{\sin x\cos x}}{\frac{\cos x\sin x+1}{\cos x\sin x}}=\frac{\cos x-\sin x}{\cos x\sin x+1}\times \frac{2}{2}=\frac{2\left( \cos x-\sin x \right)}{2\cos x\sin x+2}$

So $\int{\frac{\csc x-\sec x}{1+\sec x\csc x}dx}=\int{\frac{\cos x-\sin x}{\cos x\sin x+1}dx}=\int{\frac{2\left( \cos x-\sin x \right)}{2\cos x\sin x+1+1}dx}$

But ${{\cos }^{2}}x+{{\sin }^{2}}x=1\Leftrightarrow 2\cos x\sin x+1=2\cos x\sin x+{{\cos }^{2}}x+{{\sin }^{2}}x={{\left( \cos x+\sin x \right)}^{2}}$

So $\int{\frac{\csc x-\sec x}{1+\sec x\csc x}dx}=2\int{\frac{\cos x-\sin x}{{{\left( \cos x+\sin x \right)}^{2}}+1}dx}$

Take $u=\cos x+\sin x\Leftrightarrow du=\left( \cos x-\sin x \right)dx$

So $\int{\frac{\csc x-\sec x}{1+\sec x\csc x}dx}=2\int{\frac{du}{{{u}^{2}}+1}=2\arctan u+c=2\arctan \left( \cos x+\sin x \right)+c}$