Integral $\int\frac{\sin2x}{1+x^2}\,dx$

Question

$$\int\dfrac{\sin\;2x}{1+x^2}\;dx$$

My approach:

First of all I can try expanding $\sin \;2x$.

Which gives us,

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

Then I can assume something like(probably flawed):

Let $\sin(x)=u$, thus, $\cos(x)dx=du$. Then, the integral would become,

$$\int\dfrac{u\;}{1+(\sin^{-1}\;u)^2}\;du$$.

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I am not able to think the forward step from here. Or probably I must use other methods of integration(not sure)....Is there any mistakes in my above approach? If yes, where? And if not, what would be the forward step?

Answer 1

let $I(t)=\int_0^x \frac{sin(tx)}{1+x^2}, I'(t)=\int_0^x \frac{cos(tx)t}{1+x^2}, I''(t)=\int_0^x \frac{-sin(tx)t^2}{1+x^2}$ which yields the equation if, we use integration by parts $-I(t)t^2+\int_0^x 2tI(t)\frac{dt}{dx}dx = I''(t) -(1)$ further differentiating w.r.t t gives us $I'''(t)=-t^2I'(t)$

which means $I'(t)=A1cos(λt^2)+A2sin(λt^2)$

and I'(t) doesn't have an elementary antiderivative.

Answer 2

$$\int \frac{ \sin(2 x)}{(1 + x^2)} dx = \frac {((e^4 - 1) \,\mathrm Ci(2 i - 2 x) + (e^4 - 1) \,\mathrm (2 x + 2 i) + i (1 + e^4) (\mathrm Si(2 i - 2 x) + \mathrm Si(2 x + 2 i)))}{(4 e^2)} + C$$