Evaluate the indefinite integral
$$\int\left(a^2-\frac{\sin {(a)}}{a^{2} + 1}-a \cos {(a)}\right)\,da$$
I tried a lot of integration strategies to solve this, but I can't find what will solve the derivative. Can you give me a hint on how to solve this?
Specifically, I am thinking that the fact that
$$\int\frac{1}{x^2 + 1}\,dx = \tan^{-1} (x) + C$$
can be applied here.
Let me share my solution
$$\int\left(a^2-\frac{\sin {(a)}}{a^{2} + 1}-a \cos {(a)}\right)\,da =\int\frac{a^2(a^2-1)-\sin {(a)}-a(a^2-1) \cos {(a)}}{a^{2} + 1}\,da$$
$$=\int\frac{a^4 -a^3\cos {(a)}- a^2+a\cos {(a)}-\sin {(a)} }{a^{2} + 1}\,da$$
I got stuck here. I hope you can help me solve the problem. Thank you in advance for your comments.
$$\int\left(a^2-\frac{\sin {(a)}}{a^{2} + 1}-a\cos{(a)}\right)\,da=\int\left(a^2-a\cos{(a)}\right)\,da-\int\frac{\sin {(a)}}{a^{2} + 1}\,da$$ The first integral does not make any problem.
Considering the second $$\frac{\sin {(a)}}{a^{2} + 1}=\frac{\sin {(a)}}{(a+i)(a-i)}=\frac i 2\left( \frac{\sin {(a)}}{a+i}-\frac{\sin {(a)}}{a-i}\right)$$ Let us consider $$I_k=\int\frac{\sin {(a)}}{a+ki}\,da\qquad \text{with}\qquad k=\pm 1$$ make $b=a+ki$, replace and expand the sine function $$I_k=\int\frac{\sin {(a)}}{a+ki}\,da=\int\frac{\sin (b-i k)}{b}\,db=\int \frac{\sin (b) \cosh (k)-i \cos (b) \sinh (k)}{b}\,db$$ $$I_k=\cosh (k)\int\frac{\sin (b)}b \,db-i\sinh (k)\int\frac{\cos (b)}b \,db$$ $$I_k=\cosh (k)\,\text{Si}(b)-i\sinh (k)\,\text{Ci}(b)$$ where appear the sine and cosine exponential integral functions.