Question

помогите, пожалуйста, с интегралами

Answer 1

$1)\; \; \int \frac{dx}{cos^2x\sqrt{2tgx-1}}=\frac{1}{2}\int \frac{d(2tgx-1)}{\sqrt{2tgx-1}}=\sqrt{2tgx-1}+C\\\\2)\; \; \int sin^{10}x\, cosx\, dx=\int sin^{10}x\cdot d(sinx)=\frac{sin^{11}x}{11}+C\\\\3)\; \; \int \frac{x\, dx}{\sqrt{5-x^2}}=-\frac{1}{2}\int \frac{d(5-x^2)}{\sqrt{5-x^2}}=-\sqrt{5-x^2}+C\\\\4)\; \; \int \frac{dx}{1-5x^2}=\frac{1}{\sqrt5}\int \frac{d(\sqrt5x)}{1-(\sqrt5x)^2}=\frac{1}{\sqrt5}\cdot \frac{1}{2}\cdot ln\Big |\frac{1+\sqrt5x}{1-\sqrt5x}\Big |+C$

$5)\; \; \int \underbrace {(1-x)}_{u}\underbrace {e^{3x}\, dx}_{dv}=[\, du=-dx\; ,\; v=\frac{1}{3}e^{3x}\, ]=\\\\=\frac{1}{3}(1-x)e^{3x}+\frac{1}{3}\int e^{3x}\, dx=\frac{1}{3}(1-x)e^{3x}+\frac{1}{9}\, e^{3x} +C\\\\6)\; \; \int (\frac{4}{9x}+x^3-\frac{4}{tg^2(8x-7)})dx=\frac{4}{9}\, ln|x|+\frac{x^4}{4}+\int ctg^2(8x-7)dx=\\\\=\frac{4}{9}\, ln|x|+\frac{x^4}{4}+\int (\frac{1}{sin^2(8x-7)}-1)\, dx=\\\\=\frac{4}{9}\, ln|x|+ \frac{x^4}{4}-\frac{1}{8}\, ctg(8x-7)-x+C$