How do you integrate $int 1/theta^2cos(1/theta)$ ?

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Answer 1 · 2016-12-16T21:51:23

$-sin(1/theta) + C$

Let $u = 1/theta$ . Then $du = -1/theta^2 d theta$ and $d theta = -du(theta)^2$ .

$=int(1/theta^2 cosu *-du (theta)^2) du$

$= -int(cosu)du$

$=-sinu + C$

Resubstitute:

$=-sin(1/theta) + C$

Hopefully this helps!

How do you integrate int 1/theta^2cos(1/theta)int 1/theta^2cos(1/theta)?