Question

Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number $n$.

If $I_n=\int e^{\alpha x}{\sin }^{n}xdx=\frac{e^{\alpha x}}{{\alpha }^2+n}{\sin }^{n-1}x(\alpha \sin x-n\cos x)+A\int e^{\alpha x}{\sin }^{n-2}x\cos x$, then A is equal to

• A. $\frac{n(n-1)e^{\alpha x}}{{\alpha }^2+n^2}$
• B. $\frac{n-1}{{\alpha }^2+n^2}$
• C. $\frac{n-1(n-2)}{{\alpha }^2+n^2}$
• D. $\frac{n(n-2)}{{\alpha }^2+n^2}$

Answer 1

The correct option is A $\frac{n(n-1)e^{\alpha x}}{{\alpha }^2+n^2}$

$I_n=\int e^{\alpha x}{\sin }^{n}xdx=\frac{e^{\alpha x}}{\alpha }{\sin }^{n}x-\frac{n}{\alpha }\int e^{\alpha x}{\sin }^{n-1}x\cos xdx$

$I_n=\frac{e^{\alpha x}}{\alpha }{\sin }^{n}x-\frac{n}{{\alpha }^2}{e}^{\alpha x}{\sin }^{n-1}x\cos x+\frac{n}{{\alpha }^2}\int \left[\right(n-1){\sin }^{n-2}x{\cos }^{2}x-{\sin }^{n}x]e^{\alpha x}dx$

$I_n(1+\frac{n^2}{{\alpha }^2})=\frac{e^{\alpha x}}{{\alpha }^2}{\sin }^{n-1}x[\alpha \sin x-n\cos x]+\frac{n(n-1)}{{\alpha }^2}\int e^{\alpha x}{\sin }^{n-2}xdx$

$I_n=\frac{e^{\alpha x}}{{\alpha }^2+n^2}{\sin }^{n-1}x[\alpha \sin x-n\cos x]+\frac{n(n-1)}{{\alpha }^2+n^2}{I}_{n-2}$