MathCS.org - Real Analysis: Example 7.2.6(d): Applying Integration by Parts

Example 7.2.6(d): Applying Integration by Parts

Find sin⁵(x) dx

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As in the previous example, it seems that integration by parts does not apply because there is no product. But we can write sin⁵(x) = sin(x) sin⁴(x) and define the functions

g'(x) = sin(x) so that g(x) = -cos(x)
f(x) = sin⁴(x) so that f'(x) = 4 sin(x)³ cos(x)

Then G(x) = f(x) g(x) = -cos(x) sin⁴(x) and

sin⁵(x) dx = -cos(x) sin⁴(x) + 4 sin(x)³ cos²(x) dx =
= -cos(x) sin⁴(x) + 4 sin(x)³ (1 - sin²(x)) dx =
= -cos(x) sin⁴(x) + 4 sin(x)³ dx - 4 sin⁵(x) dx

It seems as if we are stuck. Not only did we get the original integral sin⁵(x) dx back, but we introduced sin³(x) dx which seems just about as complicated as the original integral. However, with a little more abstraction we can solve the original problem "by magic".

Let

I_k = sin^k(x) dx = sin(x) sin^k-1(x) dx

Define the functions

g'(x) = sin(x) so that g(x) = -cos(x)
f(x) = sin^k-1(x) so that f'(x) = (k-1) cos(x) sin^k-2(x)

Then

I_k = -cos(x) sin^k-1(x) + (k-1) cos² sin^k-2(x) dx =
      = -cos(x) sin^k-1(x) + (k-1) (1 - sin²) sin^k-2(x) dx =
      = -cos(x) sin^k-1(x) + (k-1) sin^k-2(x) dx - (k-1) sin^k(x) dx =
      = -cos(x) sin^k-1(x) + (k-1) I_k-2 - (k-1) I_k

Solving this expression for I_k we get

I_k = -1/k cos(x) sin^k-1(x) + (k-1)/k I_k-2

But this is a recursive formula for computing I_k for any integer k. In particular we have

I₀ = sin⁰(x) dx = 1 dx = x
I₁ = sin¹(x) dx = -cos(x)
I₂ = -1/2 cos(x) sin(x) + 1/2 I₀ = -1/2 cos(x) sin(x) + 1/2 x
I₃ = -1/3 cos(x) sin²(x) + 2/3 I₁ = -1/3 cos(x) sin²(x) - 2/3 cos(x)
I₄ = -1/4 cos(x) sin³(x) + 3/4 I₂ =
= -1/4 cos(x) sin³ + 3/4 (-1/2 cos(x) sin(x) + 1/2 x)
I₅ = -1/5 cos(x) sin⁴(x) + 4/5 I₃ =
= -1/5 cos(x) sin⁴(x) + 4/5 (-1/3 cos(x) sin²(x) - 2/3 cos(x))

Indeed, a - cumbersome - check would show that I₅ = sin⁵, so that our recursive formula and the resulting answer for our particular example seem correct.

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Example 7.2.6(d): Applying Integration by Parts