Proving $\int_{0}^{\infty} e^{-ax}\sin^{p}(x) dx$

Question

My text book says that

$$ I = \int_{0}^\infty e^{-ax} sin^p ~x ~dx = \frac {p (p – 1)}{p^2 + a^2} \int_{0}^\infty e^{-ax} sin^{p-2}~x~dx $$

Can someone help me find this result? Because I only see when differentiating by parts $$ u = sin^p (x) $$ $$ du = p~cos(x)~sin^{p-1}~x $$ $$ dv = e^{-ax} $$ $$ v = \frac{–e^{-ax}}{a}$$ $$ so~I = 0 + \frac{p}{a}\int_{0}^\infty e^{-ax} cos(x)~sin^{p-1}~x~dx $$

and then again

$$ u = cos(x) sin^{p-1} (x) $$ $$ du = -sin^p (x) + (p-1)cos^2 (x) sin^{p-2} (x) $$ $$ dv = e^{-ax} $$ $$ v = \frac{–e^{-ax}}{a}$$ $$ so~I = 0 + \frac{p}{a^2}\int_{0}^\infty e^{-ax} ((p-1)cos^2(x) sin^{p-2} (x)~–~sin^p (x)) dx $$

and from there I get no further than $$ so~I = \frac{p}{a^2}\int_{0}^\infty e^{-ax} sin^{p-2}x~((p-1)cos^2(x) – sin^2 (x)) dx $$ ?

Where am I wrong in my attempt?

Answer 1

As David said use $\cos^2x=1-\sin^2x$, which gives $$I = \dfrac{p}{a^2}\int_{0}^\infty e^{-ax} \sin^{p-2}\left((p-1) – p\sin^2(x) \right) dx$$ $$I = \dfrac{p(p-1)}{a^2}\int_{0}^\infty e^{-ax} \sin^{p-2}\, dx-\frac{p^2}{a^2}I$$ Now by finding there $I$ we will achieve required

Answer 2

Ok, I have got it now, continuing indeed with $$\cos^2 = 1 – \sin^2 :$$ $$\text{This makes}\ I = \frac{p}{a^2}\int_{0}^\infty e^{-ax} \sin^{p-2}((p – \sin^2(x) – 1 ) dx $$

which shows the desired recurrence $$I = \frac{p}{a^2}(p(I_{-2})~–~pI~–~I_{-2}) $$ $$I + \frac{p^2}{a^2}I= \frac{p^2}{a^2}(I_{-2})~–~\frac{p}{a^2}I_{-2} $$ $$I\left(\frac{a^2 + p^2}{a^2}\right) = (I_{-2})\frac{p^2 – p}{a^2}$$ $$I = (I_{-2})\frac{p^2 – p}{a^2 + p^2}$$ the comment in my textbook that this was an ‘easy’ deduction kept me from looking into this (to me) less obvious continuation … thank you for your remarks!