Question

find the integral of :
cosx/1-cos^2 dx

Answer 1

$$\int \frac {\cos x} {1 - \cos^2 x} \, \mathrm{d}x$$
From the normalization identity, we know that \(\sin^2 x = 1 - \cos^2 x\).

Answer 2

Thus:
$$\int \frac{\cos x}{\sin^2 x} \, \mathrm{d}x$$

Answer 3

Let us substitute \(u = \sin x\). Thus, \(\mathrm{d}u = \cos x \, \mathrm{d}x\).

Answer 4

Sorry about previous answers, I meant:
$$\int \frac{\mathrm{d}u}{u^2}
= -\frac 1 2 u^{-1}$$

Answer 5

Okay, let's substitute \(u = \sin x\).

Answer 6

The answer is $$-\frac{1}{2 \sin x}$$

Answer 7

thanks

Answer 8

another one please

Answer 9

???

Answer 10

Sure.

Answer 11

integral
sinx(2cscx+sinx)dx

Answer 12

$$\int (\sin x)(2 \csc x + \sin x) \, \mathrm{d}x$$

Answer 13

\(\csc x = \frac 1 {\sin x}\).

Answer 14

yes

Answer 15

ok

Answer 16

1+sin^2x

Answer 17

Kind of. You forgot the 2.
$$\int 2 + \sin^2 x \, \mathrm{d}x$$.

Answer 18

ok

Answer 19

the last one?

Answer 20

???

Answer 21

Have you got the answer to the previous integral?

Answer 22

no

Answer 23

x+...?

Answer 24

?????????????