Интегрални задачи

[shtelian]

Здравейте, необходима ми е отново вашата помощ, относно тези задачи, които прикачвам, като при наличие на а=8. Ще съм много благодарен, ако някой помогне. Поздрави

95E8C900-3A16-4F0D-8DBA-960F64663C2B.jpeg (166.94 KiB) Прегледано 940 пъти

[Добромир Глухаров]

1 зад.

$\int\frac{cosx}{1+asinx}dx=\frac{1}{a}\int\frac{acosx}{1+asinx}dx=\frac{1}{a}\int\frac{d(asinx)}{1+asinx}=\frac{1}{a}\int\frac{d(1+asinx)}{1+asinx}=\frac{1}{a}ln|1+asinx|+C$

$\int\frac{dx}{ax^2+ax+6}=\int\frac{dx}{ax^2+2\sqrt{a}x\frac{\sqrt{a}}{2}+\frac{a}{4}+6-\frac{a}{4}}=\int\frac{dx}{\left(\sqrt{a}x+\frac{\sqrt{a}}{2}\right)^2+\frac{24-a}{4}}=\frac{1}{\sqrt{a}}\int\frac{d\left(\sqrt{a}x+\frac{\sqrt{a}}{2}\right)}{\left(\sqrt{a}x+\frac{\sqrt{a}}{2}\right)^2+\frac{24-a}{4}}=\frac{1}{\sqrt{a}}\int\frac{d\frac{\sqrt{a}x+\frac{\sqrt{a}}{2}}{\frac{24-a}{4}}}{\frac{\left(\sqrt{a}x+\frac{\sqrt{a}}{2}\right)^2}{\frac{24-a}{4}}+1}=\frac{1}{\sqrt{a}\frac{\sqrt{24-a}}{2}}\int\frac{d\frac{\sqrt{a}x+\frac{\sqrt{a}}{2}}{\frac{\sqrt{24-a}}{2}}}{\left(\frac{\sqrt{a}x+\frac{\sqrt{a}}{2}}{\frac{\sqrt{24-a}}{2}}\right)^2+1}=\frac{2}{\sqrt{24a-a^2}}arctan\frac{\sqrt{a}x+\frac{\sqrt{a}}{2}}{\frac{\sqrt{24-a}}{2}}+C$

[Добромир Глухаров]

5 зад

$\int_1^\infty x^ae^{-x^a}dx$

$\underline{a\geq1}\Rightarrow x^a\geq x\Rightarrow x^ae^{-x^a}\leq xe^{-x}\Rightarrow\int_1^\infty x^ae^{-x^a}dx\leq\int_1^\infty xe^{-x}dx=-\int_1^\infty xde^{-x}=-xe^{-x}|_1^\infty+\int_1^\infty e^{-x}dx=-0+1.e^{-1}-e^{-x}|_1^\infty=\frac{1}{e}-0+\frac{1}{e}=\frac{2}{e}\\\Rightarrow$ интегралът е сходящ.

[Добромир Глухаров]

2 задача

$\int_0^1\frac{dx}{(1+ax^2)^2}=\int_0^1\frac{1+ax^2-ax^2}{(1+ax^2)^2}dx=\int_0^1\frac{dx}{1+ax^2}-\frac{1}{2}\int\frac{x.2axdx}{(1+ax^2)^2}=\int_0^1\frac{dx}{1+ax^2}-\frac{1}{2}\int\frac{xd(1+ax^2)}{(1+ax^2)^2}=\int_0^1\frac{dx}{1+ax^2}-\frac{1}{2}\int xd\frac{(1+ax^2)^{-2+1}}{-2+1}=\int_0^1\frac{dx}{1+ax^2}+\frac{1}{2}\int xd\frac{1}{1+ax^2}=\int_0^1\frac{dx}{1+ax^2}+\frac{1}{2}\left(\frac{x}{1+ax^2}|_0^1-\int_0^1\frac{dx}{1+ax^2}\right)=\frac{1}{2}\int_0^1\frac{dx}{1+ax^2}+\frac{1}{2}\cdot\frac{x}{1+ax^2}|_0^1=\frac{1}{2\sqrt{a}}\int_0^1\frac{d(\sqrt{a}x)}{1+(\sqrt{a}x)^2}+\frac{1}{2}\left(\frac{1}{1+a}-0\right)=\frac{1}{2\sqrt{a}}arctan(\sqrt{a}x)|_0^1+\frac{1}{2(1+a)}+C=\frac{arctan(\sqrt{a})}{2\sqrt{a}}-0+\frac{1}{2(1+a)}+C=\frac{arctan(\sqrt{a})}{2\sqrt{a}}+\frac{1}{2(1+a)}+C$

$\int_1^2\frac{lnx}{ax^2}dx=\frac{1}{a}\int_1^2lnxd\left(-\frac{1}{x}\right)=\frac{1}{a}lnx\left(-\frac{1}{x}\right)_1^2+\frac{1}{a}\int_1^2\frac{dlnx}{x}=-\frac{lnx}{ax}|_1^2+\frac{1}{a}\int_1^2\frac{dx}{x^2}=-\left(\frac{ln2}{2a}-\frac{ln1}{a}\right)+\frac{1}{a}\cdot\left(-\frac{1}{x}\right)|_1^2=-\frac{ln2}{2a}+\frac{0}{a}-\frac{1}{a}\left(\frac{1}{2}-1\right)+C=-\frac{ln2}{2a}+\frac{1}{2a}+C=\frac{1-ln2}{2a}+C$