Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Could anyone solve this please

Any help will be appreciated :smile:

:smile:

IMG_0090.jpg320.8KB

Original post by Freddy-Francis

Any help will be appreciated :smile:

:smile:

Split the integral like so:

\displaystyle \int \cos^n (2x) \, \mathrm{d}x = \int \cos^{n-2} (2x) \cos^2 2x \, \mathrm{d}x = \int \cos^{n-2} (2x) (1-\sin^2 2x) \, \mathrm{d}x

.

Then this is basically just:

I_n = \int \cos^{n-2} (2x) \, \mathrm{d}x - \int \sin^2 2x \cos^{n-2} 2x \, \mathrm{d}x

The latter term can be tackled via IBP with

u = \sin 2x

and

\mathrm{d}v = \sin 2x \cos^{n-2} 2x

OP

Original post by Zacken

Split the integral like so:

\displaystyle \int \cos^n (2x) \, \mathrm{d}x = \int \cos^{n-2} (2x) \cos^2 2x \, \mathrm{d}x = \int \cos^{n-2} (2x) (1-\sin^2 2x) \, \mathrm{d}x

.

Then this is basically just:

I_n = \int \cos^{n-2} (2x) \, \mathrm{d}x - \int \sin^2 2x \cos^{n-2} 2x \, \mathrm{d}x

The latter term can be tackled via IBP with

u = \sin 2x

and

\mathrm{d}v = \sin 2x \cos^{n-2} 2x

oh ok thanks :smile:

:smile:

Can you do d for me please if you have time.

Original post by Freddy-Francis

oh ok thanks :smile:

:smile:

Can you do d for me please if you have time.

I'm not going to do it for you, but think - you want to make the x^n become x^(n-1) so the obvious thing to do is to differentiate that. Furthermore, you know that e^x is its own integral and derivative. So the obvious thing to do is to use IBP with

u = x^{n}

and

\mathrm{d}v = e^x

.

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