Question

integrationsin³x.dx result​

Answer 1

Answer:

$\sf \int{sin^3x}dx$

here use sin3x formula ,

$\sf sin \: 3x = 3sinx \: - 4 {sin}^{3}x$

∴ $\sf sin 3x - 3sinx = -4sin^3x$

∴ $\sf sin^3x =(\frac{sin3x - 3sinx}{-4})$

take a (-) on numerator and denominator side of RHS

∴ $\sf sin^3x = (\frac{-sin3x + 3sinx}{4})$

∴ $\sf sin^3x = (\frac{3sinx - sin3x}{4})$ ---------(1)

$\sf \int{sin^3x}dx$

put the value of sin³x from equation (1)

$\sf = \int ({\frac{3sinx - sin3x}{4}})dx$

$\sf = \int {\frac{3sinx}{4}}dx - \int{\frac{sin3x}{4}}dx$

$\sf = \frac{3}{4}\int{sinx}dx-\frac{1}{4} \int{sin3x}dx$

$\sf=\frac{3}{4}(-cosx)-\frac{1}{4}(\frac{-cos3x}{3})+c$

$\sf =(\frac{-3cosx}{4})-(-(\frac{cos3x}{12}))+c$

$\sf =(-\frac{3cosx}{4})+(\frac{cos3x}{12})+c$

$\sf =(\frac{cos3x}{12}) -(\frac{3cosx}{4} )+ c$

$\boxed {\sf I = \frac{1}{12}cos3x - \frac{3}{4}cosx + c}$

Answer 2

Note: Check this attachment..!

So:

$\boxed{\sf{\int sin^{3}xdx\:=\:-cos\:x+\frac{1}{3}\:cos^{3}x+C}}$