Q.$\displaystyle\int_0^1 \dfrac{\sin(3\tan^{-1}x)\tan^{-1}x}{1+x^2}\,dx$

Step 1. Substitute $\theta=\tan^{-1}x$. Then $d\theta=\dfrac{dx}{1+x^2}$, and the limits become $x=0\Rightarrow\theta=0$, $x=1\Rightarrow\theta=\pi/4$. Since $\dfrac{\tan^{-1}x}{1+x^2}dx=\theta\,d\theta$, the integral becomes

Step 2. Recognize this is now a Bernoulli's-formula case. $u=\theta$ is a polynomial (degree 1) and $v=\sin3\theta$ is repeatedly integrable, so Bernoulli's formula applies with $u^{(1)}=1,\ u^{(2)}=0$.

Step 3. List the successive anti-derivatives of $v=\sin3\theta$.

$v_{(1)}=\displaystyle\int\sin3\theta\,d\theta=-\dfrac13\cos3\theta$

$v_{(2)}=\displaystyle\int v_{(1)}\,d\theta=-\dfrac19\sin3\theta$

Step 4. Apply Bernoulli's formula. $\int uv\,d\theta=uv_{(1)}-u^{(1)}v_{(2)}$ (terminates since $u^{(2)}=0$):

Step 5. Self-verify by differentiating. $F'(\theta)=-\dfrac13\cos3\theta+\theta\sin3\theta+\dfrac13\cos3\theta=\theta\sin3\theta$ ✓, matching the integrand.

Step 6. Evaluate $F(\pi/4)-F(0)$. At $\theta=\pi/4$: $\cos\tfrac{3\pi}{4}=-\tfrac{\sqrt2}2,\ \sin\tfrac{3\pi}{4}=\tfrac{\sqrt2}2$.

At $\theta=0$: $F(0)=0(\cos0)+\tfrac19\sin0=0$.

Step 7. Combine over a common denominator. $\dfrac{\pi\sqrt2}{24}=\dfrac{3\pi\sqrt2}{72}$ and $\dfrac{\sqrt2}{18}=\dfrac{4\sqrt2}{72}$, so

Substitution $\theta=\tan^{-1}x$ reduces the integrand to a polynomial-times-sine form, then Bernoulli's formula

• Trying to apply Bernoulli's formula directly to the original $x$-integral — $\tan^{-1}x$ is not a polynomial, so its 'derivatives' never terminate
• Forgetting to change the limits of integration ($0$ and $1$ in $x$ become $0$ and $\pi/4$ in $\theta$) after substituting