Integral Calculus Question 182
Question: If $ f(p,q)=\int_0^{\pi /2}{{{\cos }^{p}}x\cos qxdx} $ , then
Options:
A) $ f(p,q)=\frac{q}{p+q}f(p-1,q-1) $
B) $ f(p,q)=\frac{p}{p+q}f(p-1,q-1) $
C) $ f(p,q)=\frac{p}{p+q}f(p-1,q-1) $
D) $ f(p,q)=-\frac{q}{p+q}f(p-1,q-1) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ f(p,q)=\int_0^{\pi /2}{{{\cos }^{p}}x\cos qxdx} $ $ =[ {{\cos }^{p}}x.\frac{\sin qx}{q} ]_0^{\pi /2}+\int_0^{\pi /2}{\frac{p}{q}{{\cos }^{p-1}}x\sin x\sin qxdx} $ $ =0+\frac{p}{q}\int_0^{\pi /2}{{{\cos }^{p-1}}x[\cos (q-1)x-\cos qx\cos x]dx} $ $ [\because \cos (q-1)x=\cos qx\cos x+\sin qx\sin x $
$ \therefore \cos (q-1)x-\cos qx $ $ \cos x=\sin qx\sin x) $ $ =\frac{p}{q}f(p-1,q-1)-\frac{p}{q}f(p,q) $
$ \Rightarrow ( 1+\frac{p}{q} )f(p,q)=\frac{p}{q}f(p-1,q-1) $
$ \Rightarrow f(p,q)=\frac{p}{p+q}f(p-1,q-1) $
BETA
