Prove that
sin m x cos n x dx = – 
+ 
sin m–2 x cos n x dx
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. The smaller of the indices of sin x and cos x in the two integrals to be connected are m –2 and n respectively. So, we define
P = sin m–1 x cos n+1 x
P = sin m–1 x cos n+1 x
∴ 
= (m –1) sin m–2 x cos n+2 x –(n+1) sin m x cos n x
⇒ 
= (m –1) sin m–2 x cos n x cos 2 x – (n +1) sin m x cos n x
⇒ 
= (m –1) sin m–2 x cos n x(1 –sin 2 x) – (n+1) sin m x cos n x
⇒ 
= (m –1) sin m–2 x cos n x –(m –1) sin m x cos n x – (n + 1) sin m x cos n x
⇒ 
= (m –1) sin m–2 x cos n x – (m –1 + n + 1) sin m x cos n x
⇒ 
= (m –1) sin m–2 x cos n x – (m + n) sin m x cos n x
Integrating both sides with respect to x, we obtain
P = (m –1) 
sin m–2 x cos n x dx – (m + n) 
sin m x cos n x dx ⇒ (m + n) 
sin m x cos n x dx = – P + (m –1) 
sin m–2 x cos n x dx
⇒ 
sin m x cos n x dx = – 
+ 
sin m–2 x cos n x dx
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