Question

∫ sin2x cos 3x dx= A cosx+Bcos 5x+c,then A+B=......,Select Proper option from the given options.
(a) 1/5
(b) 3/10
(c) 3/5
(d) 2/5

Answer 1

here is your answer a

Answer 2

Given, $\int{sin2x.cos3x}\,dx=Acosx+Bcos5x+C$

we know , 2sinC.cosD = sin(C + D) + sin(C - D)
so, sin2x.co3x = 1/2[2sin2x.cos3x]
= 1/2[sin(2x + 3x) + sin(2x - 3x)]
= 1/2[ sin5x - sinx]

now,
$\int{sin2x.cos3x}\,dx=\frac{1}{2}\int{2sin2x.cos3x}\,dx\\\\\\=\frac{1}{2}\int{(sin5x-sinx)}\,dx\\\\\\=\frac{1}{2}\int{sin5x}\,dx-\frac{1}{2}\int{sinx}\,dx\\\\\\=\frac{1}{2}\frac{-cos5x}{5}+\frac{1}{2}cosx+C\\\\\\=-\frac{1}{10}cos5x+\frac{1}{2}cosx+C$

compare both equation,
we get, B = -1/10 and A = 1/2
so, A + B = 1/2 -1/10 = (5 - 1)/10 = 4/10 = 2/5
hence, option (d) is correct.