Question

Sa se calculeze următoarele integrale:
S tg^2x dx
S x^3+2/x+1 dx
S (sinx+cosx)^2 dx
S (sin2x+cos2x)^2 dx
S (tgx + ctgx)^2 dx
S 1+sin^2x/cos^2x dx
S x^4+4/x^2+1 dx
S 2^2x e^x dx

Answer 1

$\displaystyle \int tg^2 x dx=\int (tg^2 x+1)dx -\int 1 dx=tg x-x +C\\\int \dfrac{x^3+2}{x+1} dx=\int \dfrac{x^3+1}{x+1} dx+\int \dfrac{1}{x+1} dx= \int \dfrac{(x+1)(x^2-x+1)}{x+1} dx +\\+ln(x+1)=\int (x^2-x+1) dx +ln(x+1) = \dfrac{x^3}{3}-\dfrac{x^2}{2}+ x +\ln (x+1) +C\\\\\int (\sin x+\cos x)^2 dx=\int (sin^2 x+2\sin x\cos x+\cos^2 x)dx = \\\int (1+\sin 2x) dx= \int 1 dx+\int \sin 2x dx=x - \dfrac{\cos 2x}{2}+C\\\\\int (\sin 2x+\cos 2x)^2 dx= \int (\sin^2 (2x)+2\sin 2x\cos 2x+\cos^2 2x) dx=$

$\displaystyle \int (1+\sin 4x) dx=\int 1 dx +\int \sin 4x dx= x -\dfrac{\cos 4x }{4}+C\\\int (tg x+ctg x)^2 dx= \int (tg^2 x+2tg x\cdot ctg x+ctg^2 x)dx =\\\int (tg^2 x+2 +ctg^2 x) dx=\int (tg^2 x+1) dx +\int (ctg^2 x+ 1) dx = tg x-ctg x\\\int \dfrac{1+\sin^2 x}{\cos^2 x} dx=\int \dfrac{1}{\cos^2 x} dx +\int tg^2 x dx =tg x+tg x- x+C=\\=2tgx-x+C\\ \int \dfrac{x^4+4}{x^2+1} dx =\int \dfrac{(x^2+1)^2 -2x^2-2+6-1}{x^2 +1} dx=\int (x^2+1) dx - \int 2 dx\\ +5\int \dfrac{1}{x^2+1} dx =$

$\dfrac{x^3}{3}+x-2x+5arctg x+C=\dfrac{x^3}{3} -x+5arctg x+C\\\displaystyle \int 2^{2x}\cdot e^x dx=\int 4^x\cdot e^x dx=\int (4e)^x dx = \dfrac{(4e)^x}{\ln(4e)}+C=\dfrac{(4e)^x}{\ln 4+1}+C$