How do you take the anti-derivative of sin^5t without using identities?

Question

When I say identities I mean trig identities. Would the anti-derivative be (1/6)cos^6(t)

Answer 1

it is not what you specified

It cannot be solved without identities

sin ^5 t = sin^4 t sin t
= (1-cos^2 t)^2 sin t

now let cost = x
so diffentiate to get - sint dt = dx

so we need to integrate

-(1-x^2)^2 dx

= -dx + 2 x^2 dx - x^4 dx

integrating we get

-x + 2x^3/3 - x^5/5 + C
= - cos t + 2 cos^3 t/3 - cos^5 t /5 + C

where C is constant of integration

Answer 2

the only way without using identities is to integrate from first principles

Answer 3

without using identities? you need the indentity that the integral of sint is -cost....without that i dont know how to help you

Answer 4

I think you can do it if you don't consider the exponential form of sin to be an identity. sin(t) = (e^(i*t) - e^(-i*t) )/(2i)

Integral of sin(t)^5 = Integral of ((e^(i*t) - e^(-i*t) )/(2i))^5

You can carry it from here.

Hope this help

Answer 5

try integration by parts