Answer to Question #136211 in Calculus for qwerty

Question

\int \:\frac{x+sinx}{1+cosx}dx

Answer 1

$I=\int \:\frac{x+\sin x}{1+\cos x}dx={\displaystyle\int}\left(\dfrac{\sin\left(x\right)}{\cos\left(x\right)+1}+\dfrac{x}{\cos\left(x\right)+1}\right)\mathrm{d}x$

Thus,

Applying linearity we get,

$I={\displaystyle\int}\dfrac{\sin\left(x\right)}{\cos\left(x\right)+1}\,\mathrm{d}x+{\displaystyle\int}\dfrac{x}{\cos\left(x\right)+1}\,\mathrm{d}x$

Now, consider

$I_1={\displaystyle\int}\dfrac{\sin\left(x\right)}{\cos\left(x\right)+1}\,\mathrm{d}x$

Let,

$u=1+\cos x\implies \sin x dx =-du$

, hence

$I_1=-{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u=-\ln u=-\ln (1+\cos x)+c_1$

Consider,

$I_2={\displaystyle\int}\dfrac{x}{\cos\left(x\right)+1}\,\mathrm{d}x$

Thus, after rewriting the above integrand we get

$I_2={\displaystyle\int}\dfrac{x\csc\left(x\right)}{\csc\left(x\right)+\cot\left(x\right)}\,\mathrm{d}x$

Now, integrate by parts, take $f=x$ , $g'=\dfrac{\csc\left(x\right)}{\csc\left(x\right)+\cot\left(x\right)}$

And apply

${\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g}$

Thus, $f'=1$ and

$g={\displaystyle\int}\dfrac{\csc\left(x\right)}{\csc\left(x\right)+\cot\left(x\right)}\,\mathrm{d}x\\$

Now, substitute $u=\dfrac{1}{\csc\left(x\right)+\cot\left(x\right)}$ ,hence $\mathrm{d}x=-\dfrac{\left(\csc\left(x\right)+\cot\left(x\right)\right)^2}{-\csc^2\left(x\right)-\cot\left(x\right)\csc\left(x\right)}\,\mathrm{d}u$

Therefore,

$g=\int du=u\implies g=\dfrac{1}{\csc\left(x\right)+\cot\left(x\right)}$

Hence,

$I_2=\dfrac{x}{\csc\left(x\right)+\cot\left(x\right)}-{\displaystyle\int}\dfrac{1}{\csc\left(x\right)+\cot\left(x\right)}\,\mathrm{d}x$

Now, consider

$I_3={\displaystyle\int}\dfrac{1}{\csc\left(x\right)+\cot\left(x\right)}\,\mathrm{d}x$

After simplification of $I_3$ we get

$I_3=-{\displaystyle\int}-\dfrac{1}{\cos\left(x\right)+1}{\sin x} dx$

Take

$v=1+\cos x\implies -\sin xdx=dv$

Hence,

$I_3=-\int \frac{1}{v}dv=-\ln v=-\ln(1+\cos x)+c_2$

Thus,

$I_2=\dfrac{x}{\csc\left(x\right)+\cot\left(x\right)}-I_3\\ \implies I_2=\dfrac{x}{\csc\left(x\right)+\cot\left(x\right)}+\ln (1+\cos x)+c_2$

Therefore,

$I=I_1+I_2\implies I=\dfrac{x}{\csc\left(x\right)+\cot\left(x\right)}+c$

Where, constant $c=c_1+c_2$