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$$I=\int_{0}^{1}\frac{x^2e^{\arctan{x}}}{\sqrt{x^2+1}}\,dx $$ First, i noticed that $$ I=\int_{0}^{1} xe^{\arctan{x}}\sin(\arctan{x})\,dx.$$ Using the substitution $\arctan{x}=t$, we get that$$I=\int_{0}^{1}\frac{\sin^2{t}}{\cos^3{t}}e^t\, dt=I_1-I_2,$$ where $$I_1=\int_{0}^{\frac{\pi}{4}}\frac{e^t}{\cos^3{t}}\, dt,$$ and $$I_2=\int_{0}^{\frac{\pi}{4}}\frac{e^t}{\cos{t}}\,dt$$ How could i continue?

Jack
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  • why not use the substitution on the original integral? if I am not completely wrong, you end up with something very close to a Gaussian function – lmaosome Aug 31 '21 at 06:38
  • @lmaosome, you re right, i could ve used the substitution from the very beginning and get the same result. For the Gaussian function, it is the first time i heard about it =) – Jack Aug 31 '21 at 06:52

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Since the derivative of $\sec t$ is $\sec t \tan t$ we need to play with this.

We have $\frac{\sin^2{t}}{\cos^3{t}}e^t = e^t \tan^2t \sec t = \frac12 e^t \sec t (\tan^2 t + \tan^2 t) = \frac12 e^t \sec t (\sec^2 t + \tan^2 t -1) = \frac12(e^t \sec^3t + e^t \sec t \tan^2t - e^t \sec t ) = \frac12(e^t\sec^3t + e^t \sec t \tan^2t + e^t \sec t \tan t - e^t \sec t - e^t \sec t \tan t)$

Now we see that $e^t \sec t + e^t \tan t \sec t = (e^t \sec t)' $

and $e^t\sec^3t + e^t \sec t \tan^2t + e^t \sec t \tan t = e^t \sec t (\tan t)' + e^t \tan t (\sec t)' + (e^t)' \sec t \tan t = (e^t \tan t\sec t)'$

So we have $e^t \tan^2 t \sec t = \frac12 ( e^t \sec t \tan t - e^t \sec t )'$

So $ \int_0 ^ {\pi/4} e^t \tan^2 t \sec t \,dt = \frac12 e^0 \sec (0) = \frac 12$

Infinity_hunter
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