I have this difficult integral to solve.
$$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 – 1} dx} $$
Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^{\sqrt 2 – 1}$ as $(\sin x)^{\sqrt 2}.(\sin x)$ and $(\sin x)^{\sqrt 2 – 2}.(\sin x)$ respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please!
Edit:
This is what I did (showing just for the numerator)
$$ \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx $$
$$ = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx $$
$$ = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 – 1 }(\cos^2 x) dx $$
(taking $ v = \sin x $ and $ u = (\sin x)^{\sqrt 2} $ in the $ \int uv $ formula)
$$ = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 – 1 } – (\sin x)^{ \sqrt 2 + 1 } \right) dx $$
Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer.
Best Answer
Let the top integral be $I$, and the bottom one $J$. To make typing simpler, let $a=\sqrt{2}$.
Integrate the top one by parts, letting $du=\sin x$ and $v=\sin^{a} x$. This is the standard way to get a reduction formula for $\int \sin^n x\,dx$.
So $dv=a\cos x\sin^{a-1}x\,dx$ and we can take $u=-\cos x$. Then $$I=\left. -\cos x \sin^{a}x\right|_0^{\pi/2}+\int_0^{\pi/2}a\cos^2 x\sin^{a-1} x\,dx.$$ The first part dies at both ends. Rewrite $\cos^2 x$ as $1-\sin^2 x$. Then $$I=aJ -aI.$$ Now we get $$I=\frac{a}{a+1}J$$ and it's over.