#int (1+sinx)/(sinx * (1+cosx))dx# ?

Question

Isabella Ackerman · Accepted Answer

#1/4(tan(x/2))^2+tan(x/2)+1/2Ln(tan(x/2))+C# #int (1+sinx)/(sinx*(1+cosx))*dx# After using #y=tan(x/2)#, #dx=(2dy)/(y^2+1)#, #sinx=(2y)/(y^2+1)# and #cosx=(1-y^2)/(y^2+1)# transforms, this integral became #int (1+(2y)/(y^2+1))/((2y)/(y^2+1)(1+(1-y^2)/(y^2+1)))*(2dy)/(y^2+1)# =#int ((y^2+2y+1)/(y^2+1))/((2y)/(y^2+1)2/(y^2+1))*(2dy)/(y^2+1)# =#int ((y^2+2y+1)*dy)/(2y)# =#1/2int y*dy+int dy+1/2int (dy)/y# =#y^2/4+y+1/2Lny+C# =#1/4(tan(x/2))^2+tan(x/2)+1/2Ln(tan(x/2))+C#

Emma Aldridge · Answer

undefined To integrate the given expression $\frac{{1+\sin(x)}}{{\sin(x) \cdot (1+\cos(x))}}$ with respect to $x$, you can use trigonometric identities and substitution.

Let $u = 1 + \cos(x)$, then $du = -\sin(x) dx$.

Substitute $u = 1 + \cos(x)$ and $du = -\sin(x) dx$ into the integral:

$\int \frac{{1 + \sin(x)}}{{\sin(x) \cdot (1 + \cos(x))}} dx$

$-\int \frac{{1 + \sin(x)}}{{u}} du$

$-\int \left(1 + \sin(x)\right) \cdot \frac{1}{{u}} du$

$-\int \left(1 + \sin(x)\right) \cdot \frac{1}{{1 + \cos(x)}} du$

$-\int \frac{{1}}{{1 + \cos(x)}} du - \int \frac{{\sin(x)}}{{1 + \cos(x)}} du$

$=-\ln|1 + \cos(x)| - \int \frac{{\sin(x)}}{{1 + \cos(x)}} du$

For the second integral, let $v = 1 + \cos(x)$, then $dv = -\sin(x) dx$.

So, the second integral becomes:

$-\int \frac{{\sin(x)}}{{1 + \cos(x)}} du$

$-\int \frac{{\sin(x)}}{{v}} dv$

$-\int \frac{1}{v} dv$

$=-\ln|1 + \cos(x)| + C$

Thus, the integral of $\frac{{1+\sin(x)}}{{\sin(x) \cdot (1+\cos(x))}}$ with respect to $x$ is $-\ln|1 + \cos(x)| + C$, where $C$ is the constant of integration.

Adam Adkins · Answer

undefined To integrate the function $\frac{{1 + \sin x}}{{\sin x (1 + \cos x)}}$ with respect to $x$, you can first express it as the sum of two fractions:

$$
\frac{{1 + \sin x}}{{\sin x (1 + \cos x)}} = \frac{1}{{\sin x (1 + \cos x)}} + \frac{{\sin x}}{{\sin x (1 + \cos x)}}
$$

Then, integrate each term separately.

For the first term:
$$
\int \frac{1}{{\sin x (1 + \cos x)}} \, dx
$$
You can use the substitution method. Let $u = 1 + \cos x$, then $du = -\sin x \, dx$.

Substituting $du$ and solving for the integral, you get:
$$
\int \frac{1}{u} \, du = \ln|u| + C = \ln|1 + \cos x| + C_1
$$

For the second term:
$$
\int \frac{{\sin x}}{{\sin x (1 + \cos x)}} \, dx = \int \frac{1}{{1 + \cos x}} \, dx
$$

Again, you can use substitution. Let $v = \tan\left(\frac{x}{2}\right)$, then $dx = \frac{2}{1 + v^2} \, dv$.

Substituting $dx$ and solving for the integral, you get:
$$
\int \frac{1}{{1 + \cos x}} \, dx = \int \frac{2}{1 + v^2} \, dv = 2 \arctan(v) + C = 2 \arctan\left(\tan\left(\frac{x}{2}\right)\right) + C_2
$$

Finally, combining both results:
$$
\int \frac{{1 + \sin x}}{{\sin x (1 + \cos x)}} \, dx = \ln|1 + \cos x| + 2 \arctan\left(\tan\left(\frac{x}{2}\right)\right) + C
$$