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Iet \( I_{n, m}=\int \sin ^{n} x \cos ^{m} x d x \). Then, we can relate \( I_{n, m} \) with each of the following :
(i) \( I_{n-2, m} \)
(ii) \( I_{n+2, m} \)
(iii) \( I_{n, m-2} \)
(iv) \( I_{n, m+2} \)
(v) \( I_{n-2, m+2} \)
(vi) \( I_{n+2, m-2} \)
Suppose we want to establish a relation between \( I_{n, m} \) and \( I_{a, m-2} \), then we get
\[
P(x)=\sin ^{n+1} x \cos ^{m-1} x
\]
In \( I_{n, m} \) and \( I_{n, m-2} \) the exponent of \( \cos x \) is \( m \) and \( m-2 \)
respectively, the minimum of the two is \( m-2 \) adding 1 to the
minimum we get \( m-2+1=m-1 \). Now, choose the exponent
\( m-1 \) of \( \cos x \) in \( P(x) \) Similarly, choose the exponent of \( \sin x \) for
\[
P(x)=(n H) \sin ^{n} x \cos ^{m} x-(m-1) \sin ^{n+2} x \cos ^{m-2} x
\]
Now, differentiating both the sides of Eq. (i), we get
\[
\begin{array}{l}
=(n+1) \sin ^{n} x \cos ^{m} x-(m-1) \sin ^{n} x\left(1-\cos ^{2} x\right) \cos ^{m-2} x \\
=(n+1) \sin ^{n} x \cos ^{m} x-(m-1) \sin ^{n} x \cos ^{m-2} x \\
\quad+(m-1) \sin ^{n} x \cos ^{n} x \\
=(n+m) \sin ^{n} x \cos ^{m} x-(m-1) \sin ^{n} x \cos ^{m-2} x
\end{array}
\]
Now, integrating both the sides, we get
\[
\sin ^{n+1} x \cos ^{m-1} x=(n+m) I_{n, m}-(m-1) I_{n, m-2}
\]
Similarly, we can establish the other relations.
The relation between \( I_{4,2} \) and \( I_{2,2} \) is
(a) \( I_{4,2}=\frac{1}{6}\left(-\sin ^{3} x \cos ^{3} x+3 I_{2,2}\right) \)
(b) \( I_{4,2}=\frac{1}{6}\left(\sin ^{3} x \cos ^{3} x+3 I_{2,2}\right) \)
(c) \( I_{4,2}=\frac{1}{6}\left(\sin ^{3} x \cos ^{3} x-3 I_{2,2}\right) \)
(d) \( I_{4,2}=\frac{1}{4}\left(-\sin ^{3} x \cos ^{3} x+2 I_{2,2}\right) \)
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