integration of cos x/(1+sin x)^1/2?

Answer 1

∫cos x / (1+sin x)^1/2 dx = log(cos(x/2) + sin (x/2)) + c

Good luck.

Answer 2

You need a integral of f(u) * du

Let's make (1 + sin x ) = u
then cos x = du

So... You really have u^(-1/2) * du

That integral is not so bad!
2*u^(+1/2)

or 2*(1+sin x)^(1/2)

Differentiate our answer to see if it gives you the original problem!

Answer 3

for this we can do u substitution. We let u=(1+sin x). This means du= cos x. Remember the derivative of a constant is zero.

So we get du/(u)^1/2 or u^(-1/2) du.

The integral of the above statement is 2*u^(1/2). Then we replace the u with the orginal statement. So our answer is

2*(1+sin x)^(1/2)

Answer 4

Rewrite it first:

Integral [cos (x) * (1 + sin (x) )^ -(1/2) * dx]

Then substitute:

u = 1 + sin (x)

Derive:

du = cos (x) * dx
or
dx = du / cos (x)

Now, substitute:

Integral [cos (x) * (1 + sin (x) )^ -(1/2) * dx]
=
Integral [cos (x) * (u)^ -(1/2) * (du / cos (x)]

and re-arrange:
Integral [cos (x) * (u)^ -(1/2) * (du / cos (x)]
=
Integral [ [cos (x) / cos (x)] * [(u)^ -(1/2)] * [du] ]
=
Integral [ [(u)^ -(1/2)] * [du] ]


Use the reverse power rule:
Integral [ [(u)^ -(1/2)] * [du] ]
=
2 * u^(1/2) + C

And re-substitute

Integral [cos (x) * (1 + sin (x) )^ -(1/2) * dx]
=
2 * [1 + sin (x)]^(1/2) + C

Answer 5

= vital( sin x dx) + vital (sin x cos x dx) For comparing the 2d time period, sin x cos x dx = a million/2 * 2 sin x cos x dx = a million/2 sin 2x dx enable 2x = t. Then dt = 2 dx So it truly is going to develop into a million/4 sin t dt Integrating, it truly is going to develop into, -a million/4 cos t = -a million/4 cos 2x So very last answer is -cos x - a million/4 cos 2x