Question
If $$f\left( {p,\,q} \right) = \int_0^{\frac{\pi }{2}} {{{\cos }^p}x\,\cos \,qx\,dx,} $$ then :
A.
$$f\left( {p,\,q} \right) = \frac{q}{{p + q}}f\left( {p - 1,\,q - 1} \right)$$
B.
$$f\left( {p,\,q} \right) = \frac{p}{{p + q}}f\left( {p - 1,\,q - 1} \right)$$
C.
$$f\left( {p,\,q} \right) = - \frac{p}{{p + q}}f\left( {p - 1,\,q - 1} \right)$$
D.
$$f\left( {p,\,q} \right) = - \frac{q}{{p + q}}f\left( {p - 1,\,q - 1} \right)$$
Answer :
$$f\left( {p,\,q} \right) = \frac{p}{{p + q}}f\left( {p - 1,\,q - 1} \right)$$
Solution :
$$\eqalign{
& f\left( {p,\,q} \right) = \int_0^{\frac{\pi }{2}} {{{\cos }^p}x\,\cos \,qx\,dx} \cr
& \Rightarrow f\left( {p,\,q} \right) = \left[ {{{\cos }^p}x.\frac{{\sin \,qx}}{q}} \right]_0^{\frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {\frac{p}{q}{{\cos }^{p - 1}}x\,\sin \,x\,\sin \,qx\,dx} \cr
& \Rightarrow f\left( {p,\,q} \right) = 0 + \frac{p}{q}\int_0^{\frac{\pi }{2}} {{{\cos }^{p - 1}}} x\left[ {\cos \left( {q - 1} \right)x - \cos \,qx\,\cos \,x} \right]dx \cr} $$
\[\left[ \begin{gathered}
\because \,\cos \left( {q - 1} \right)x = \cos \,qx\,\cos \,x + \sin \,qx\,\sin \,x \hfill \\
\therefore \,\cos \left( {q - 1} \right)x - \cos \,qx\,\cos \,x = \sin \,qx\,\sin \,x \hfill \\
\end{gathered} \right]\]
$$\eqalign{
& \Rightarrow f\left( {p,\,q} \right) = \frac{p}{q}f\left( {p - 1,\,q - 1} \right) - \frac{p}{q}f\left( {p,\,q} \right) \cr
& \Rightarrow \left( {1 + \frac{p}{q}} \right)f\left( {p,\,q} \right) = \frac{p}{q}f\left( {p - 1,\,q - 1} \right) \cr
& \Rightarrow f\left( {p,\,q} \right) = \frac{p}{{p + q}}f\left( {p - 1,\,q - 1} \right) \cr} $$