Solve this integral...?

Question

Can someone please help me solve this integral for Math 222.

the integral of 1/(1+sinx) dx

Thank you so much!

Answer 1

Integral ( (1 / [1 + sin(x)] dx)

This is obviously difficult to integrate in its current form. Let's try multiplying top and bottom by the conjugate of the bottom; that is,
1 - sin(x). Note that by doing so, we would be changing the denominator into a difference of squares.

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

Now, apply the trig identity to the denominator.

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

Now, separate this into two fractions.

Integral ( (1/cos^2(x)) - sin(x)/cos^2(x) ) dx

By definition, sec(x) = 1/cos(x), so we have

Integral ( sec^2(x) - sin(x)/cos^2(x) ) dx

And now, we split this into two integrals.

Integral ( sec^2(x) ) dx - Integral (sin(x)/cos^2(x) ) dx

The first integral is easy; it's one of our known derivatives, and is equal to tan(x). The second integral we can solve by substitution.

Let u = cos(x). Then
du = -sin(x) dx, and
(-1)du = sin(x) dx, so we have

tan(x) - Integral ( 1/u^2 (-du) )

Pulling out the (-1) out of the integral, we get

tan(x) + Integral (1/u^2) du

And 1/u^2 is just u^(-2), and integrate this give us (-1)u^(-1), or
-1/u

tan(x) - 1/u + C

But u = cos(x), so

tan(x) - 1/cos(x) + C

tan(x) - sec(x) + C

Answer 2

Don't know what's up with that y^5 up there, but here goes your integral... int y^4 from 0 to 2 = y^5/5 2^5/5 - 0^5/5 32/5 - 0 integral = 32/5 I can't explain here how to solve integrals. There are many different methods for many different types of integrals. Yours is the simplest, but explaining why and how would take a long, long time.

Answer 3

The trick for this integral is to multiply both the numerator and denominator by (1-sinx). This gives the integral of (1-sinx)/(1-(sinx)^2). 1-(sinx)^2 = (cosx)^2, so the integrand is now (1-sinx)/(cosx)^2. Now separate the integral into two integrals. The first is the integral of 1/(cosx)^2, which is the integral of (secx)^2. This integral is just tanx. The second one is the integral of -sinx/(cosx)^2. Now just do a u-substitution with u = cosx, du = -sinxdx, and the integral is simple.

Answer 4

Integrate 1/(1 + sinx) with respect to x.

First rearrange the terms to make it easier to integrate.

1/(1 + sinx) = (1 - sinx)/{(1 - sinx)(1 + sinx)}
= (1 - sinx)/(1 - sin²x) = (1 - sinx)/cos²x
= 1/cos² - sinx/cos²x = sec²x - (tanx)(secx)

Now we can integrate.

∫{1/(1 + sinx)}dx = ∫{sec²x - (tanx)(secx)}dx = tanx - secx + C

Answer 5

http://www.wolframalpha.com/input/?i=integral
used this can help a lot =)

Answer 6

Us solving it wont allow you to learn anything.