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 $c=c_1+c_2$