Mathematics

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

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


ANSWER

$$\displaystyle \frac {n(n - 1)e^{\alpha x}}{\alpha^2 + n^2}$$


SOLUTION
$$I_{n}=\displaystyle\int e^{\alpha x}\sin^{n} x d{x}=\dfrac{e^{\alpha x}}{\alpha}\sin^{n} x-\dfrac{n}{\alpha}\int e^{\alpha x}\sin^{n-1}x\cos x d {x}$$
$$I_{n}=\dfrac{e^{\alpha x}}{\alpha}\sin^{n}x-\dfrac{n}{\alpha^2}e^{\alpha x}\sin^{n-1}x\cos x+\dfrac{n}{\alpha ^2}\displaystyle\int [(n-1)\sin ^{n-2}x\cos^2 x-\sin^{n}x]e^{\alpha x} d{x}$$
$$I_{n}\bigg(1+\dfrac{n^2}{\alpha^2}\bigg)=\dfrac{e^{\alpha x}}{\alpha^2}\sin^{n-1}x [\alpha \sin x-n\cos x]+\dfrac{n(n-1)}{\alpha^2}\displaystyle\int e^{\alpha x}\sin^{n-2}x d{x}$$
$$I_{n}=\dfrac{e^{\alpha x}}{\alpha^2+n^2}\sin^{n-1}x[\alpha \sin x-n\cos x]+\dfrac{n(n-1)}{\alpha^2+n^2}I_{n-2}$$
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Single Correct Hard Published on 17th 09, 2020
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