Integrating powers of sin x

How do you find

I_n = ∫ sinⁿ x dx

This is easy and satisfying to do using Integration by parts:

∫ u dv = uv - ∫ v du

Take u=sin ^n-1x and dv = sin x dx

This means v = -cos x, and du=(n-1) cos x sin ^n-2x dx

Then I_n = ∫ u dv = -cos x sin^n-1x + (n-1)∫ cos²x sin^n-2x dx

Substituting cos²x = 1 - sin²x inside the integral sign, notice that we get an iterative relationship involving I_n-2

Then I_n = -cos x sin^n-1x + (n-1)(I_n-2 - I_n)

nI_n = -cos x sin^n-1x + (n-1)I_n-2 -mathformula not MarkupType-
Fatal error: Uncaught Error: Call to undefined method mathformula::makeTypeDef() in /var/www/vhosts/ent.storkey.uk/html/inc/reporting2.php:365 Stack trace: #0 /var/www/vhosts/ent.storkey.uk/html/inc/dynaPage.php(189): echo_value('intsinpower', 'content', Array) #1 /var/www/vhosts/ent.storkey.uk/html/inc/dynaPage.php(742): dynaPage->echo_value('content') #2 /var/www/vhosts/ent.storkey.uk/html/inc/dynaPage.php(179): mainPanel->write('main-panel') #3 /var/www/vhosts/ent.storkey.uk/html/index.php(21): dynaPage->write() #4 {main} thrown in /var/www/vhosts/ent.storkey.uk/html/inc/reporting2.php on line 365