Интегралы!!! помогите

$a)\int\limits^2_1 {(4x+3-\frac{4}{x^2})} \, dx=\frac{4x^2}{2}+3x-\frac{4*x^{-1}}{-1}=(2x^2+3x+\frac{4}{x})|\limits^2_1=\\=(8+6-1)-(2+3+4)=4$

$b)\int\limits^4_1{(\frac{\sqrt{x}}{x}+8(2x-5)^3)dx}=\int\limits^4_1{(\frac{\sqrt{x}}{\sqrt{x}\sqrt{x}}+8(2x-5)^3)dx}=\\=\int\limits^4_1{(\frac{1}{\sqrt{x}}+8(2x-5)^3)dx}=\frac{x^\frac{1}{2}}{\frac{1}{2}}+\frac{8(2x-5)^4}{4*2}=\\=(2\sqrt{x}+(2x-5)^4)|\limits^4_1=(2*2+81)-(2+81)=\\=4+81-2-81=2$

$c)\int\limits^{-\frac{\pi}{4}}_{-\frac{\pi}{2}}{\frac{dx}{cos^2x-1}} = \int\limits^{-\frac{\pi}{4}}_{-\frac{\pi}{2}}{\frac{dx}{1-sin^2x-1}} = \int\limits^{-\frac{\pi}{4}}_{-\frac{\pi}{2}}{-\frac{dx}{sin^2x}} = -(-ctgx)=\\=ctgx|\limits^{-\frac{\pi}{4}}_{-\frac{\pi}{2}}=ctg(-\frac{\pi}{4})-ctg(-\frac{\pi}{2})=-1-0=-1$

$d)\int\limits^{2\pi}_0{(cos\frac{x}{8}-sin\frac{x}{8})^2}dx=d)\int\limits^{2\pi}_0{(cos^2\frac{x}{8}-2cos\frac{x}{8}sin\frac{x}{8}+sin^2\frac{x}{8})}dx=\\ =\int\limits^{2\pi}_0{(1-sin\frac{x}{4})}dx=(x+4cos\frac{x}{4})|\limits^{2\pi}_0=(2\pi+4cos\frac{\pi}{2})-\\-(0+4cos0)=2\pi-4$