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

dx√2ax − x 2 = a n sin – 1 xa − 1 ⋅The value of n is

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

n the interval [0, π], graphs of sin x, 2 sin x and sin 2x are _______ x - axesand some portion of the graph of sin 2x lies _______ x-axes

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

Q 60. Find dy/dx, given y = (2tanx * sin2x) / (sec'x - 1) ? Ops: A. •-8cosxsinx B. 0-8cos-x C. © -4cosxsinx D. 0-8cos2x.sin2