∫ dx√2ax − x 2 = a n sin – 1 xa − 1

dx√2ax − x 2 = a n sin – 1 xa − 1 ⋅The value of n is(a) 0 (b) –1 (c) 1 (d) none of these.You may use dimensional analysis to solve the problem

Let x and a stand for distance. Is ∫ dx√a 2 − x 2= 1a sin − 1 ax dimensionally correct ?

a Let k ∈ N. Find an expression in terms of k for ∫ π20 sin2k x dx

∫? x√x+5dx

1∫01x2 + √x dx