Evaluation of $$\int \cot x\left(\csc x-1\right)e^xdx = $$

$\bf{My\; Try::}$ Let $$I = \int e^x \cot x\cdot \csc xdx-\int e^x \cot xdx$$

Using Integration by parts, We get

$$ I = -e^x\cdot \csc x+\int e^x \csc x-e^x \cdot \cot xdx $$

So $$I = -e^x\cdot \csc x+\int \left(\csc x-\cot x\right)e^x dx$$

So $$I = -e^x\cdot \csc x+\int \frac{1-\cos x}{\sin x}e^x dx$$

So $$I = -e^x\cdot \csc x+\int e^ x\cdot \tan \frac{x}{2}dx$$

So $$I = -e^x\cdot \csc x+2e^ x\cdot \ln \sec \frac{x}{2}+2\int e^x \cdot \ln \cos \frac{x}{2}dx$$

Now how can I calculate integration of $$\int e^x\cdot \ln \cos \frac{x}{2}dx$$

Although We can not Express in terms of elementary form.

Can we express it using Series Expansion, If yes Then plz explain here.


  • 2
    $\begingroup$ Neither W.A. nor Maple produce a closed-form expression for an antiderivative. $\endgroup$ – Travis Jan 17 '16 at 11:09
  • $\begingroup$ Your anti-derivative possesses a closed form if and only if $y'+y=\cot x~(\csc x-1)$ does also. $\endgroup$ – Lucian Jan 17 '16 at 11:13
  • $\begingroup$ Thanks Travis and Lucian, Can we solve it Using Series expansion, If yes Then plz explain here. $\endgroup$ – juantheron Jan 17 '16 at 11:30

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