Home page for accesible maths Math 101 Chapter 5: Integration

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

5.20 Wallis’s integral

Example

To evaluate In=∫0π/2cosn⁡x⁢d⁢xI_{n}=\int_{0}^{\pi/2}\cos^{n}x\,dx for integers n≠0.n\neq 0.

Solution. First we find:

I0=∫0π/2cos0⁡x⁢d⁢x=π/2;I_{0}=\int_{0}^{\pi/2}\cos^{0}x\,dx=\pi/2;
I1=∫0π/2cos⁡x⁢d⁢x=[sin⁡x]0π/2=sin⁡(π/2)-sin⁡0=1.I_{1}=\int_{0}^{\pi/2}\cos x\,dx=\bigl[\sin x\bigr]_{0}^{\pi/2}=\sin(\pi/2)-% \sin 0=1.

We shall prove the recurrence relation

In=n-1nIn-2  (n≥2);I_{n}={{n-1}\over{n}}I_{n-2}\qquad(n\geq 2);

note that we go down two steps from nn to n-2n-2. We integrate by parts, obtaining

Example of integrating by guesswork

Find ∫cos2⁡x⁢sin⁡x⁢d⁢x\int\cos^{2}x\sin x\,dx.