Question

Решите неопределенные интегралы:

Answer 1

$1)\; \int \frac{sin^4x}{cos^2x}dx=\int \frac{sin^2x\cdot sin^2x}{cos^2x}dx=\int tg^2x\cdot sin^2x\, dx=\\\\=\int tg^2x\cdot (1-cos^2x)dx=\int tg^2x\, dx-\int tg^2x\cdot cos^2x\, dx=\int \frac{sin^2x}{cos^2x}dx-\\\\-\int \frac{sin^2x}{cos^2x}\cdot cos^2x\, dx=\int \frac{1-cos^2x}{cos^2x}dx-\int sin^2x\, dx=\int \frac{dx}{cos^2x}-\int dx-\\\\-\int \frac{1-cos2x}{2}dx=tgx-x-\frac{1}{2}(x-\frac{1}{2}sin2x)+C$

$2)\; \int \frac{dx}{\sqrt[3]{x}+2\sqrt[4]{x}}=[\, x=t^{12},\, dx=12t^{11}dt,\; \sqrt[3]{x}=\sqrt[3]{t^{12}}=t^4,\, \sqrt[4]{x}=t^3\, ]=\\\\=\int \frac{12\cdot t^{11}\, dt}{t^4+2t^3}=12\int \frac{t^3\cdot t^8}{t^3(t+2)}dt=12\int \frac{t^8}{t+2}dt=\\\\=12\int (t^7-2t^6+4t^5-8t^4+16t^3-32t^2+64t-128-\frac{256}{t+2})dt=\\\\=12(\frac{t^8}{8}-\frac{2t^7}{7}+\frac{2t^6}{3}-\frac{8t^5}{5}+4t^4-\frac{32t^3}{3}+32t^2-128t-\\\\-256\cdot ln|t+2|+C,\; \; gde\; \; t=\sqrt[12]{x}$