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Pattern - 31 of Indefinite Integration

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⋆ Pattern – 31

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

$I_n=\ \frac{-\sin^{n-1}{x}\times\cos{x}}{n}+\ \frac{n-1}{n}\ I_{n-2}\ :n\geq2$

Here, $I_n$ = ∫ $\sin^n{x}$ dx

∴ $I_n$ = ∫ $\sin^{n-1}{x}$ × sin x dx

= $\sin^{n-1}{x}\ sin\int{\sin{x}\ dx}$ – $\ \int\left(\frac{d}{dx}\sin^{n-1}{x\ \int{\sin{x}\ dx}}\right)dx$

= – $\sin^{n-1}{x}$ ∙ cos x + ∫ (n-1) $\sin^{n-2}{x}$ × cos x cos x dx

We know that,

∫ uv dx = u ∫ vdx – ∫ $\left(\frac{du}{dx}\ \int v\ dx\right)dx$ with the use of LIATE 1

$= -\sin^{n-1}{x}\cos{x}+\left(n-1\right)\int\sin^{n-2}{\left(1-\sin^2{x}\right)}dx$

$= -\sin^{n-1}{x} \cos{x}$ + $\ \left(n-1\right)\ \int\sin^{n-2}{x}\ dx$ – $\ \left(n-1\right)\ \int\sin^n{x}\ dx$

= $\ -\sin^{n-1}{x}\cos{x}$ + $\left(n-1\right)I_{n-2\ }$ – $\left(n-1\right)I_n$

∴ (1+n-1) I_n = $-\sin^{n-1}{x}\times \cos{x}+\left(n-1\right)I_{n-2}$

∴ n = – $\sin^{n-1}{x}\times \cos{x\ }+\left(\ n-1\right)\ I_{n-2}$

∴ = $\frac{-\sin^{n-1}{x}\ \times\ \cos{x}}{n}+\ \frac{n-1}{n}\ I_{n-2}\ : n \geq 2$

1) Evaluate : ∫ $\sin^6{x}$ dx

Solution. Let,

$I_6=\ \int\sin^6{x}\ dx$

$=\ -\ \frac{\sin^5{x}\times\cos{x}}{6}+\frac{5}{6}\ I_4$

$=\ \frac{-\sin^5{x}\times\cos{x}}{6}+\frac{5}{6}\ \left[\frac{-\sin^3{x}\times\cos{x}}{4}+\frac{3}{4}\ I_2\right]$

$=\ \frac{-\sin^5{x}\times\cos{x}}{6}-\ \frac{5\sin^3{x}\times\cos{x}}{24}+\ \frac{5}{8}\left[\frac{-\sin{x\cos{x}}}{2}+\frac{1}{2}I_0\right]$

$=\ \frac{-\sin^5{x}\cos{x}}{6}-\ \frac{5\sin^3{x}}{24}-\ \frac{5}{16}\sin{x}\times\cos{x}+\frac{5}{16}\ \int{1\ dx}$

$=\ \frac{-\sin^5{x}\cos{x}}{6}-\ \frac{5\sin^3{x}\times\cos{x}}{24}\ -\frac{5}{16}\sin{x}\times\cos{x}+\frac{4}{16}x+ c$

= ∫ $\sin^6{x} dx$