Engineering Mathematics: Multiple Sine Integration

MSI.ID.1: Recurrence Relation

$I_{n}= \int \sin ^{n}x\, dx$

(1)

$I_{n}= \int \sin ^{n}x\, dx = -\frac{1}{n}\sin ^{n-1}x\cos x + \frac{n-1}{n}I_{n-2}$

(2)

Proof

$I_{n}= \int \sin ^{n}x\, dx = \int \sin ^{n-2}x (1-\cos^{2}x)\, dx$

(3)

$I_{n}= \int \sin ^{n}x \, dx = \int \sin ^{n-2}x \, dx - \int \sin ^{n-2}x \cos^{2}x\, dx$

(4)

$I_{n}= \int \sin ^{n}x \, dx = \int \sin ^{n-2}x \, dx - \int \sin ^{n-2}x \cos x\cos x\, dx$

(5)

$u= \cos x \, , \, du= -\sin x \, dx$

(6)

$dv= \sin^{n-2}x \, dx \, \cos x \, ,\, v= \frac{\sin^{n-1}x}{n-1}$

(7)

$I_{n}= \int \sin ^{n}x \, dx = \int \sin ^{n-2}x \, dx - \frac{1}{n-1}\sin ^{n-1}\cos x- \int \frac{\sin^{n-1}x}{n-1} \, dx$

(8)

$\int \sin ^{n}x \, dx + \frac{1}{n-1}\int \sin^{n-1}x \, dx = \int \sin ^{n-2}x \, dx - \frac{1}{n-1}\sin ^{n-1}\cos x$

(9)

$\frac{n}{n-1}\int \sin^{n-1}x \, dx = \int \sin ^{n-2}x \, dx - \frac{1}{n-1}\sin ^{n-1}\cos x$

(10)

$\int \sin^{n-1}x \, dx = \frac{n}{n-1}\int \sin ^{n-2}x \, dx - \frac{1}{n}\sin ^{n-1}\cos x$

(11)

From (11) and (1)

$I_{n}= \int \sin ^{n}x\, dx = -\frac{1}{n}\sin ^{n-1}x\cos x + \frac{n-1}{n}I_{n-2}$

Engineering Mathematics