MaxWong

\(I=\int x \cos(x) dx\\ \begin{array}{rl}u=\cos x & du=-\sin x dx\\ dv=xdx&v=\dfrac{x^2}{2}\end{array}\)

\(\text{Then }\)

\(\begin{array}{rll}I=\int udv & u\cdot v = \dfrac{x^2\cos x}{2}&\int vdu = \int\dfrac{-x^2\sin x}{2}dx\\\end{array}\)

And I am stuck??

What if I am stuck in this situation? Do I do integration by parts again?

1,025,109.8 words?
Do you mean 10,251,098 words? Or that would be nonsense

\((x^{4y})^3\cdot2y^1\)

\(= x^{12y}\cdot 2y\)

\(=2x^{12y}y\)

\(\text{Also, thanks heureka and Alan for answering.}\)

\(\mbox{What if I got something like } du=2\sin x dx?\)

\(\begin{array}{rll}&&5\\-&&2\\&&3\end{array}\)

XDDD

5 - 2 = 3 .......

np :D

\(\quad 15kg \div \dfrac{1}{6}\\ = 15kg \times 6 \\ = 90kg\)

His space suit and life support packs weigh 90kg on earth.