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Parin Sir > Mathematical Proofs > Indefinite Integration > Pattern - 10 of Indefinite Integration > Pattern - 10 of Indefinite Integration

Pattern - 10 of Indefinite Integration

February 18, 2021
Category: Pattern - 10 of Indefinite Integration

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⋆ Pattern – 10

∫ $\sin^n{x}\times \cos^m{x} dx$

$\rightarrow$ If both n & m are odd, then assume for the basic of highest power = t

1) I = ∫ $\sin^3{x}\times \ \cos^7{x}$ dx

Let, cos x = t $\sin^2{x}$

∴ – sin x dx = dt $=\ \left(1-\cos^2{x}\right)$

∴ sin x dx = -dt $=1-t^2$

$\therefore\ \ \ I=\ \int\sin^2{x}\times\cos^7{x}\times\sin{x\ }dx$

$=\ -\int{\ t^7}\ \left(1-t^2\right)\ dt$

$\therefore\ \ I=\ -\ \int\left(t^7-t^9\right)\ dt$

$=\ \int\left(t^9-t^7\right)\ dt$

$=\ \frac{t^{10}}{10}-\ \frac{t^8}{8\ }+c$

$=\ \frac{\cos^{10}{x}}{10}-\ \frac{\cos^8{x}}{8}+c$

$\sin^n{x}\times \cos^m{x} dx$

$\rightarrow$ One of m & n is even & other is odd, then assume base of even power = t

2) I = ∫ $\sin^2{x}\times \cos^3{x}\ dx$

Let, sin x = t $\cos^2{x}=1-\sin^2{x}$

∴ cos x dx = dt $=1-t^2$

$\therefore\ \ I=\ \int\sin^4{x}\times\cos^2{x}\times\cos{x}\ dx$

= ∫ $t^4\ \left(1-t^2\right)\ dt$

= ∫ $\left(t^4-t^6\right)\ dt$

$\therefore\ \ I=\ \frac{t^5}{5}-\ \frac{t^7}{7}+c$

$=\ \frac{\sin^5{x}}{5}-\ \frac{\sin^7{x}}{7\ }+c$

∫ $\sin^n{x}\times \cos^m{x}\ dx$

$\rightarrow$ If both m & n are even, then reduce the index of sin x & cos x to 1 using trigonometric formulas & then integral it.

3) I = ∫ $\sin^4{x}\times \cos^2{x}\ dx$

Here,

$\sin^4{x}\times \cos^2{x}=\sin^2{x}\times \cos^2{x}\times\sin^2{x}$

$=\frac{1}{4}\ \left(2\sin{x}\times\ \cos{x}\right)^2\times\frac{1-\cos{2x}}{2}$

$=\frac{1}{8}\ \left(\sin^2{2x}\right)\ \left(1-\cos{2x}\right)$

$=\frac{1}{8}\ \left(\frac{1-\cos{4x}}{2}\right)\ \left(1-\cos{2x}\right)$

$=\frac{1}{16}\ \left(1-\cos{4x}\right)\left(1-\cos{2x}\right)$

$=\frac{1}{16}\ \left(1-\cos{2x}-\cos{4x}+\cos{4x}\ \times \cos{2x}\right)$

$=\frac{1}{32}\ \left(2-2\cos{2x}-2\cos{4x}+2\cos{4x}\times \cos{2x}\right)$

$=\frac{1}{32}\ \left(2-\cos{2x}-2\cos{4x}+\cos{6x}\right)$

$\therefore\ \ I=\frac{1}{32}\ \int\left(2-\cos{2x}-2\cos{4x}+\cos{6x}\right)dx$

$=\frac{1}{32}\ \left[2x-\ \frac{\sin{2x}}{2}-\ \frac{2\sin{4x}}{4}+\ \frac{\sin{6x}}{6}\right]+c$

$=\frac{1}{32}\times\frac{1}{6} [ 12x - 3 sin 2x - 3 sin 4x + sin 6x ] + c$

$=\ \frac{1}{192} [ 12x - 3 sin 2x - 3 sin 4x + sin 6x ] + c$

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