5 caleczek do policzenia

[BoBi]

5 calek ktore musialbym miec na jutro policzone a nie moge sobie z nimi dac rady
\(\displaystyle{ \int\sin^{4}x}\)
\(\displaystyle{ \int x^{2}(2-5x^{3})^{\frac{2}{3}}}\)
\(\displaystyle{ \int \sin(\ln x)}\)
\(\displaystyle{ \int \frac{dx}{x^{2}-x+2}}\)
\(\displaystyle{ \int \frac{x^{2}+1}{x\sqrt{x^{4}+1}}}\)

[soku11]

1)
\(\displaystyle{ \int\sin^{4}(x) =\int (1-cos^2(x))sin^2(x)dx=
t sin^{2}(x)dx-\int cos^2(x)sin^2(x)dx=
t sin^{2}(x)dx-\int \frac{(2cos(x)sin(x))^2}{4}dx=
t sin^{2}(x)dx-\frac{1}{4}\int sin^2(2x)dx=...}\)



2)
\(\displaystyle{ \int x^{2}(2-5x^{3})^{\frac{2}{3}}dx\\
2-5x^3=t\\
-15x^2dx=dt\\
x^2dx=-\frac{1}{15}dt\\
-\frac{1}{15}\int t^{\frac{2}{3}}dt=-\frac{1}{15}\cdot \frac{t^{\frac{5}{3}}}{\frac{5}{3}} +C=-\frac{t^{\frac{5}{3}}}{25}+C=...}\)



3)
\(\displaystyle{ \int \frac{x sin(ln x) }{x}dx\\
ln(x)=t\\
\frac{1}{x}dx=dt\\
x=e^t\\
\\
t e^t\cdot sin(t)dt\\
u=e^t\quad dv=sin(t)dt\\
du=e^t dt\quad v=-cos(t)\\
-e^t cos(t)+\int e^t\cdot cos(t)dt\\
u=e^t\quad dv=cos(t)dt\\
du=e^t dt\quad v=sin(t)\\
-e^t cos(t)+e^t sin(t) - t e^t\cdot sin(t)dt\\
\\
t e^t\cdot sin(t)dt=-e^t cos(t)+e^t sin(t) - t e^t\cdot sin(t)dt\\
2\int e^t\cdot sin(t)dt=-e^t cos(t)+e^t sin(t) \\
t e^t\cdot sin(t)dt=\frac{-e^t cos(t)+e^t sin(t)}{2}=...\\}\)



4)
\(\displaystyle{ \int \frac{dx}{x^2-x+2}=\int \frac{dx}{(x-2)(x+1)}=\int ft( \frac{1}{3(x-2)} -\frac{1}{3(x+1)} \right)dx=
\frac{1}{3}\int \frac{dx}{x-2} -\frac{1}{3}\int\frac{dx}{x+1}=\frac{1}{3}(ln|x-2|-ln|x+1|)=
\frac{1}{3}ln|\frac{x-2}{x+1}|+C}\)



5)
\(\displaystyle{ \int \frac{x^2+1}{x\sqrt{x^4+1}}dx=
t \frac{x^2}{x\sqrt{x^4+1}}dx+\int \frac{dx}{x\sqrt{x^4+1}}=
t \frac{x}{\sqrt{x^4+1}}dx+\int \frac{dx}{x\sqrt{x^4+1}}=...}\)
Podstawiasz \(\displaystyle{ x^2=t}\) i kombinujesz



BTW. Wpisuj, np

Kod: Zaznacz cały

frac{dx}{x}

a nie frec
POZDRO