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a)\(\int_{}^{} arcsin^2x dx\)

b) \(\int_{}^{} x(arctgx)^2dx\)

c)\(\int_{}^{} x^2cos^2x dx\)

d)\(\int_{}^{} x2^xdx\)

kakapipe pisze:a)\(\int_{}^{} arcsin^2x dx\)

\(\displaystyle \int arcsin^2x dx= \begin{bmatrix}\arcsin x=t\\x=\sin t \\ \frac{dx}{dt}=\cos t\\dx=\cos t dt \end{bmatrix}= \int t^2 \cos t dt = \int t^2( \sin t)' dt=\\
\displaystyle t^2 \sin t - 2\int t \sin t dt = t^2 \sin t + 2\int t( \cos t)' dt = t^2 \sin t + 2 t \cos t -2\int \cos t dt=\\
\displaystyle t^2 \sin t + 2 t \cos t -2 \sin t +C\)

A resztę ktoś ogarnia? : D

Arcusa dziś poskromimy, wszelką trudność w mig zwalczymy:

\(\int xarctg^2 x dx = \frac{1}{2}\int(x^2)' arctg^2 x dx = \frac{x^2arctg^2x}{2}-\int\frac{x^2}{x^2+1} \cdot arctgxdx = \frac{x^2arctg^2x}{2}-\int arctgxdx + \int \frac{arctgx}{x^2+1}dx=\\= \begin{bmatrix} t=arctgx\\dx=(x^2+1)dt\end{bmatrix} = \frac{x^2arctg^2x}{2}-\int x'arctgxdx + \int tdt =\frac{x^2arctg^2x - 2xarctgx + arctg^2x}{2}+\int \frac{xdx}{x^2+1} = \\ =\begin{bmatrix} u=x^2+1\\dx=\frac{1}{2x}du\end{bmatrix} = \frac{x^2arctg^2x - 2xarctgx + arctg^2x}{2}+\frac{1}{2}\int \frac{du}{u} =\frac{x^2arctg^2x - 2xarctgx + arctg^2x+ ln(x^2+1)}{2} + C\\\)

Choć są tego plusy i minusy, to niestraszne nam są cosinusy:

\(\int x^2 cos^2xdx = \int x^2 \cdot \frac{cos2x+1}{2}dx = \frac{1}{2}\int x^2 cos2xdx + \frac{1}{2}\int x^2dx = \begin{bmatrix} t=2x\\dx = \frac{dt}{2}\end{bmatrix}= \frac{x^3}{6}+\frac{1}{16}\int t^2costdt =\\= \frac{x^3}{6}+\frac{1}{16}\int t^2(sint)'dt = \frac{x^3}{6}+ \frac{x^2sin2x}{4} +\frac{1}{8}\int tsintdt = \frac{x^3}{6}+ \frac{x^2sin2x}{4} -\frac{1}{8}\int t(cost)'dt = \\ = \frac{x^3}{6}+ \frac{x^2sin2x - xcos2x}{4} + \frac{1}{8}\int costdt = \frac{x^3}{6}+ \frac{x^2sin2x - xcos2x}{4}+ \frac{sin2x}{8} + C\)

Nie obawiaj się przyjacielu, trudnych całek, jak nas wielu:

\(\int x2^xdx = \frac{1}{ln2}\int x (2^x)'dx = \frac{x2^x}{ln2} - \frac{1}{ln2}\int 2^xdx = \frac{x2^x}{ln2}- \frac{2^x}{ln^22}\)