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.