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

Question

Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$

My Attempt: $$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$ $$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$ $$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$ $$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$

How do I continue?

Answer 1

The integration of

$$\frac{\sin x}{1+\cos x}$$ which is of the form $f'(x)/f(x)$ is immediate. The rest is

$$\int\frac{dx}{1+\cos x}=\int\frac{dx}{2\cos^2\frac x2}$$ and

$$I=-\log|1+\cos x|+\tan\frac x2+C.$$

Answer 2

Second method is to split numerator

$$\int\frac{\sec^2(x/2)}{2}dx+\int\frac{\sin x}{1+\cos x}dx$$

So answer is $$\tan(x/2)-\ln|1+\cos x|+c$$

Answer 3

An alternate way is this $$\int \dfrac {1+\sin (x)}{1+\cos (x)} dx=\int \dfrac {1+\sin (x)}{1+\cos (x)} \dfrac {1-\cos (x)}{1-\cos (x)} dx=\int \dfrac {(1+\sin (x))(1-\cos(x))}{\sin^2(x)} dx\\=\int \csc^2(x)+\csc(x)-\csc(x)\cot(x)-\cot(x) dx$$

Answer 4

• OP's question: how to continue?
• existing answers: start over with simpler steps

So I feel the need to respond to the original question.

\begin{align} \int \frac {1+\sin (x)}{1+\cos (x)} dx &=\frac {1}{2} \int (\tan (\frac {x}{2}) +1)^2 dx \\ &= \frac12 \int (\tan^2(\frac x2)+1) dx + \int \tan(\frac x2) dx \\ &= \frac12 \int \sec^2(\frac x2) dx + \int \tan(\frac x2) dx \\ &= \tan(\frac x2) + 2\ln|\sec \frac x2| + C \end{align}

Answer 5

You were almost there, just substitute $t=\tan(\frac x2)$ now.

$\displaystyle\dfrac 12\int \left(1+\tan(\dfrac x2)\right)^2\mathop{dx}=\int \dfrac{(1+t)^2}2\times\dfrac{2\mathop{dt}}{1+t^2}=\int \left(1+\dfrac{2t}{1+t^2}\right)\mathop{dt}=t+\ln(1+t^2)+C$

Remark that you can substitute directly without intermediate trigonometric changes :

$\dfrac{1+\sin(x)}{1+\cos(x)}=\dfrac{1+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}=\dfrac{1+2t+t^2}{2}$

Answer 6

Here is another trick

$$ \begin{align} I&=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx\\ &=\int \dfrac {1+ \sin(0)+\sin (x)}{\cos(0)+\cos (x)} dx\\ &=\int \dfrac {1+ 2\sin(x/2)\cos (x/2)}{2\cos^2(x/2)}dx\\ &=\int \dfrac {1}{2\cos^2(x/2)}dx+\int \dfrac {\sin(x/2)}{\cos(x/2)}dx\\ &=\int \dfrac {1}{2\cos^2(x/2)}dx+\int {\tan(x/2)}dx\\ I&=\tan(x/2)-2\ln|\cos(x/2)|+K \end{align} $$

Answer 7

I may contribute by multiplying $(1-\cos(x))/(1-\cos(x)) $ : $$ \int \frac{1 + \sin(x)}{1 + \cos(x)} dx= \int \frac{1 + \sin(x) - \cos(x) - \sin(x)\cos(x)}{1 - \cos^{2}(x)} dx = \int \frac{1 + \sin(x) - \cos(x) - \sin(x)\cos(x)}{\sin^{2}(x)} dx = \int \frac{\sin^{2}(x) + \cos^{2}(x) + \sin(x) - \cos(x) - \sin(x)\cos(x)}{\sin^{2}(x)} dx $$ $$ = \int \left(1 + \cot^{2}(x) + \frac{1}{\sin(x)}\right) dx - \int \frac{d(\sin(x))}{\sin^{2}(x)} - \ln | \sin(x)|$$ $$= x + \int \cot^{2}(x)dx + \int \frac{1}{\sin(x)}dx + \frac{1}{\sin(x)}-\ln|\sin(x)| $$ Now for one integral, multiply by $(1+\cos(x))/(1+\cos(x)) $ : $$ \int \frac{1}{sin(x)} dx = \int \frac{1+\cos(x)}{\sin(x)(1+\cos(x))} dx = \int \frac{\sin^{2}(x) + \cos^{2}(x) + \cos(x)}{\sin(x)(1+\cos(x))} dx $$ $$ \int \frac{\sin(x)}{1+\cos(x)}dx + \int \frac{\cos(x)}{\sin(x)} dx = - \ln|1+\cos(x)| + \ln|\sin(x)| $$ so ...

$$ \int \frac{1 + \sin(x)}{1 + \cos(x)} dx= x - \ln|1+\cos(x)| + \frac{1}{\sin(x)} + \int \cot^{2}(x)dx$$ The last part may be continued..

Answer 8

From where you are you did a great job, note that

$$(\tan (\frac{x}{2}))' =\frac{1}{2}(1+\tan^2 (\frac{x}{2}))~~~and~~~ -2(\ln|\cos (\frac{x}{2})|)' = \tan (\frac{x}{2})$$

Therefore $$\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx = \dfrac {1}{2}\int (\tan^2 (\dfrac {x}{2}) +1) dx + \int \tan (\dfrac {x}{2}) dx \\=\tan (\frac{x}{2}) -2\ln|\cos (\frac{x}{2})|+c$$