Question

Integral Question: This is trignometric substitution right?

Answer 1

ok?

Answer 2

\[\int\limits_{?}^{?} 4-x/\sqrt{4-x^2}\]

Answer 3

yes

Answer 4

yeh trig substitution

Answer 5

lulu

Answer 6

my teacher posted it in the the first test section for my final exam study sheet but it doesn't look like a test 1 question

Answer 7

Saifoooooo xD

Answer 8

:D

Answer 9

No you don't need trig substitution

u = 4-x^2
du = -2x dx --> dx = -du/2x

the x's cancel, leaving
\[-\frac{1}{2}\int\limits_{}^{}4 - \frac{1}{\sqrt{u}} du\]

Answer 10

wait actually it should be
\[\int\limits_{}^{}4 + \frac{1}{2\sqrt{u}}du\]