Помогите решить 1 и 2 варианты.Спасибо!

1. [latex] \int\limits^2_1 {5} \, dx =(5x)|^{^2}_{_1}=5\\\\ \int\limits^3_1 {x^3+2x} \, dx =({1\over4}x^4+x^2)|^{^3}_{_1}=28\\\\ \int\limits^{e^2}_1 {2dx\over x} =(2\ln x)|^{^{e^2}}_{_1}=4\\\\ \int\limits^{\pi\over2}_0 {cos2x} \, dx ={1\over2}\int\limits^{\pi\over2}_0 {cos2x} \, d(2x) = {1\over2}(sin2x)|^{^{\pi\over2}}_{_0}=0\\\\ \int\limits^{\ln2}_0 {e^{-2x}} \, dx = -{1\over2}\int\limits^{\ln2}_0 {e^{-2x}} \, d(-2x) =-{1\over2}(e^{-2x})|^{^{\ln2}}_{_0}={3\over8}[/latex] [latex]\int\limits^{\pi\over4}_0 {sin{x\over2}cos{x\over2}} \, dx ={1\over2}\int\limits^{\pi\over4}_0 {sinx} \, dx =-{1\over2}(cosx)|^{^{\pi\over4}}_{_0}=-{1\over2}+{1\over2\sqrt2}[/latex] 2. [latex] \int\limits^2_{-1}{2} \, dx =(2x)|^{^2}_{_{-1}}=6\\\\ \int\limits^0_{-1} {x^2+2x} \, dx =({1\over3}x^3+x^2)|^{^0}_{_{-1}}=-{2\over3}\\\\ \int\limits^{4}_1 {3dx\over x} =(3\ln x)|^{^{4}}_{_1}=3\ln4\\\\ \int\limits^{\pi}_0 {sin5x} \, dx ={1\over5}\int\limits^{\pi}_0 {sin5x} \, d(5x) = -{1\over5}(cos5x)|^{^{\pi}}_{_0}=0.4\\\\ \int\limits^{\ln2}_{\ln1} {e^{2x}} \, dx = {1\over2}\int\limits^{\ln2}_{\ln1} {e^{2x}} \, d(2x) ={1\over2}(e^{2x})|^{^{\ln2}}_{_{\ln1}}=1.5[/latex] [latex]\int\limits^{\pi}_0 ({sin^2{x}-cos^2{x}}) \, dx =-\int\limits^{\pi}_0 {cos2x} \, dx =-{1\over2}\int\limits^{\pi}_0 {cos2x} \, d(2x) =\\-{1\over2}(sin2x)|^{^{\pi}}_{_0}=0[/latex]