Primitive function

Evaluate the following indefinite integral on $(-\frac{\pi }{2};\frac{\pi }{2})$. $$\int (\frac{1}{\cos x}-\sin x\cdot \mathrm{tg}x)\mathrm{d}x$$

Choose an incorrect evaluation of the following indefinite integral on $(0;\mathrm{\infty })$. $$\int (\sqrt{x}-1)(\sqrt{x}+1)\mathrm{d}x$$

Four girls evaluated the integral $I=\int \sin x\cdot \cos x\mathrm{d}x$ on $\mathbb{R}$. Ann started to integrate by parts like this: $I=\int \sin x\cdot \cos x\mathrm{d}x={\sin }^{2}x-\int \cos x\cdot \sin x\mathrm{d}x$. Beth started to integrate by parts like this: $I=\int \sin x\cdot \cos x\mathrm{d}x=-{\cos }^{2}x-\int \sin x\cdot \cos x\mathrm{d}x$. Claire used substitution $a=\sin x$ like this: $I=\int \sin x\cdot \cos x\mathrm{d}x=\int a\mathrm{d}a$. Diana integrated $\int \sin x\cdot \cos x\mathrm{d}x=-\cos x\cdot \sin x+c$, $c\in \mathbb{R}$. Which of the girls made a mistake?

Solve an indefinite integral $\int \frac{8x^7-30x^5}{x^8-5x^6+2}\mathrm{d}x$ on $(3;\mathrm{\infty })$.