Find #int \ (sin2x+1)/(1-sin2x) \ dx #?

Question

Emily Ammerman · Accepted Answer

# int \ (sin2x+1)/(1-sin2x) \ dx = - x -2/(tanx-1) + C #

We seek: # I = int \ (sin2x+1)/(1-sin2x) \ dx # We can write this as: # I = int \ (1+sin2x)/(1-sin2x) \ dx # # \ \ \ = -int \ (-1-sin2x)/(1-sin2x) \ dx # # \ \ \ = -int \ (1-sin2x-2)/(1-sin2x) \ dx # # \ \ \ = -int \ 1 -2/(1-sin2x) \ dx # # \ \ \ = - int \ dx + 2 \ int 1/(1-sin2x) \ dx # ..... [A] The first integral is standard, and for the second we have to perform some further analysis: Using a standard Weierstraß trigonometric identity we have: # \ \ \ \ \ \ \ sin x = (2tan(x/2))/(1+tan^2(x/2)) # # :. sin 2x = (2tan(x))/(1+tan^2(x)) = 2tanx/sec^2x # So, for the second integral, #I_2#, say,we have using this identity: # I_2 = int \ 1/(1-sin2x) \ dx # # \ \ \ = int \ 1/(1-2tanx/sec^2x) \ dx # # \ \ \ = int \ 1/( (sec^2x-2tanx)/sec^2x) \ dx # # \ \ \ = int \ (sec^2x)/( sec^2x-2tanx) \ dx # # \ \ \ = int \ (sec^2x)/( tan^2x+1-2tanx) \ dx # # \ \ \ = int \ (sec^2x)/( tanx-1 )^2 \ dx # We can now easily integrate this if we perform a substitution: Let #u= tanx-1 => (du)/dx = sec^2x # Substituting into #I_2# we have: # I_2 = int \ 1/u^2 \ du # # \ \ \ = -1/u # # \ \ \ = -1/(tanx-1) # Hence, we can now integrate [A] to get: # I = - x + 2(-1/(tanx-1)) + C # # \ \ = - x -2/(tanx-1) + C #

Evelyn Agard · Answer

undefined To find the integral of $\frac{\sin(2x) + 1}{1 - \sin(2x)}$ with respect to $x$, you can use the substitution method. Let $u = \sin(2x)$, then $du = 2\cos(2x)dx$. Rewriting $\sin(2x)$ in terms of $u$, we get:

$$\sin(2x) = u$$

Now, rewrite the integral using $u$:

$$\int \frac{1 + u}{1 - u} \cdot \frac{1}{2} du$$

This simplifies to:

$$\frac{1}{2} \int \frac{1 + u}{1 - u} du$$

Now, split the fraction:

$$\frac{1}{2} \int \left(1 + \frac{2u}{1 - u}\right) du$$

This integrates to:

$$\frac{1}{2} \left(u + 2\ln|1 - u|\right) + C$$

Replace $u$ with $\sin(2x)$ and simplify:

$$\frac{1}{2} \left(\sin(2x) + 2\ln|1 - \sin(2x)|\right) + C$$

So, the integral of $\frac{\sin(2x) + 1}{1 - \sin(2x)}$ with respect to $x$ is:

$$\frac{1}{2} \left(\sin(2x) + 2\ln|1 - \sin(2x)|\right) + C$$