Question

найти интеграл dx/(sin^2x*cos^4x) (замена t=tgx)
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Answer 1

$\int \dfrac{dx}{sin^2x\cdot cos^4x}=\int \dfrac{1}{sin^2x}\cdot \dfrac{1}{cos^2x}\cdot \dfrac{dx}{cos^2x}=\int (1+ctg^2x)\cdot (1+tg^2x)\cdot \dfrac{dx}{cos^2x}=\\\\\\=\int \Big(1+\dfrac{1}{tg^2x}\Big)(1+tg^2x)\cdot d(tgx)=\Big[\; t=tgx\; \Big]=\int \dfrac{(1+t^2)^2}{t^2}\, dt=\\\\\\=\int \dfrac{1+2t^2+t^4}{t^2}\, dt=\int \Big(\dfrac{1}{t^2}+2+t^2}\Big)\, dt=-\dfrac{1}{t}+2t+\dfrac{t^3}{3}+C=\\\\\\=-\dfrac{1}{tgx}+2tgx+\dfrac{tg^3x}{3}+C$