If I n = ∫ 0 π 2 x n sin x d x , where n > 1 , then

Question

If I n = &#8747; 0 &#960; 2 x n sin x d x , where n &#62; 1 , then

Aicharya · Accepted Answer

We have, I n = &#8747; 0 &#960; 2 x n sin x d x &#8658; I n = - x n cos x 0 &#960; 2 + &#8747; 0 &#960; 2 n x n - 1 cos x d x &#8658; I n = n &#8747; 0 &#960; 2 x n - 1 cos x d x &#8658; I n = n x n - 1 sin x 0 &#960; 2 - &#8747; 0 &#960; 2 n - 1 x n - 2 sin x d x &#8658; I n = n &#960; 2 n - 1 - n - 1 I n - 2 Hence, I n + n n - 1 I n - 2 = n &#960; 2 n - 1

If I n = &#8747; 0 &#960; 2 x n sin x d x , where n &#62; 1 , then

Solution

If I n = ∫ 0 π 2 x n sin x d x , where n > 1 , then