Question

If I_{m, n} = $\int_{}^{}\cos^{m}$x sin nx dx, then 7I_{4, 3} – 4I_{3, 2} =

• A. Constant
• B. –cos² x + c
• C. – cos⁴ x cos3x + c
• D. cos 7x – cos4x + c

Answer 1

Answer: (C)

– cos⁴ x cos3x + c

Answer 2

Integrating by parts we have

I_{4, 3} = – $\frac{\cos 3x.\cos^{4}x}{3}$ – $\frac{4}{3}$ $\int_{}^{}{\cos^{3}x.\sin x.\cos 3xdx}$

But sinx. Cos 3x = – sin2x + sin 3x. cos x.

So I_{4, 3} = – $\frac{\cos 3x.\cos^{4}x}{4}$ + $\frac{4}{3}$

$\int_{}^{}\cos^{3}x.\sin 2xdx$ – $\frac{4}{3}$ $\int_{}^{}{\cos^{4}x}.\sin 3xdx$ + C

= – $\frac{cox3x.\cos^{4}x}{3}$ + $\frac{4}{3}$ I_{3, 2} – $\frac{4}{3}$ I_{4, 3} + C

Therefore, $\frac{7}{3}$ I_{4, 3} – $\frac{4}{3}$ I_{3, 2} = – $\frac{\cos 3x.\cos^{4}x}{3}$ + C Or 7I_{4, 3} – 4I_{3, 2} = – cos 3x. cos⁴x + C