Question

If solution of $\int x^3\sin \left(\ln x^2\right)dx$ is of the form $\frac{x^A}{B}\left[A\sin \right(\ln x^2)+D\cos (\ln x^2\left)\right]+C$ (where $C$ is constant of integration), then $\frac{B}{A}-D$ is

Answer 1

$I=\int x^3\sin \left(\ln x^2\right)dx$
Put $\ln x=t$
$\Rightarrow x=e^t$
$\Rightarrow dx=e^{t}dt$
$I=\int \left(e^t{)}^3\sin \right(2t)e^{t}dt=\int \left(e^t{)}^3\sin \right(2t)e^{t}dt=\int e^{4t}\left(\sin 2t\right)dt=\frac{e^{4t}}{4^2+2^2}[4\sin 2t-2\cos 2t]+C[\because \int e^{ax}\sin bx=\frac{1}{a^2+b^2}{e}^{ax}[a\sin bx-b\cos bx]+C]\Rightarrow I=\frac{(e^{\ln x}{)}^4}{20}\left[4\sin \right(2\ln x)-2\cos (2\ln x\left)\right]+C\Rightarrow I=\frac{x^4}{20}\left[4\sin \right(\ln x^2)-2\cos (\ln x^2\left)\right]+C$
On comparing, we have
$A=4,B=20,D=-2\therefore \frac{B}{A}-D=\frac{20}{4}-(-2)=5+2=7$