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

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

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

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

Youif sintheta+costheta=m and sec theta+cosec theta=n , then n(m+1)(m-1) =

sinn x dx = − 1n sinn−1 x · cos x + n − 1n∫sinn−2 x dx. (6.4)If we take dv = sin x dx, then we have v = − cos x and we may integrate by parts withu = sinn−1 x, du = (n − 1) sinn−2 x · cos x.Using the fact that sin2 x + cos2 x = 1, one may thus conclude that∫sinn x dx = − sinn−1 x · cos x + (n − 1)∫sinn−2 x · cos2 x dx= − sinn−1 x · cos x + (n − 1)∫sinn−2 x · (1 − sin2 x) dx= − sinn−1 x · cos x + (n − 1)∫sinn−2 x dx + (1 − n)∫sinn x dx.Here, the rightmost integral coincides with the original integral on the left. Once we nowrearrange terms, we end up with n copies of the integral and equation (6.4) follows. Example 6.11 We use a reduction formula to compute the integral I3 in the case thatIn =∫xne2x dx.If we take u = xn and dv = e2x dx, then du = nxn−1 dx and v = 12 e2x, so one hasIn = 12 xne2x − n2∫xn−1e2x dx = 12 xne2x − n2 · In−1. (6.5)We now apply the last formula repeatedly to determine I3. According to the formula,I3 = 12 x3e2x − 32 · I2 = 12 x3e2x − 32 ·[12 x2e2x − I1]= 12 x3e2x − 32 ·[12 x2e2x − 12 xe2x + 12 · I0]= 12 x3e2x − 34 x2e2x + 34 xe2x − 34∫e2x dx= 12 x3e2x − 34 x2e2x + 34 xe2x − 38 e2x + C.