C4, Integerate Sin3xCos2x

Checked answer and it was simplified down to

sin3xcos2x

1/2(sin5x + sinx)

= 1/2(-1/5cos5x - cosx) + c

= -1/10(cos5x +5cosx) + c


I used all the Cos2x identities and it still won't simplify to that answer.

Checked answer and it was simplified down to

sin3xcos2x

1/2(sin5x + sinx)

= 1/2(-1/5cos5x - cosx) + c

= -1/10(cos5x +5cosx) + c


I used all the Cos2x identities and it still won't simplify to that answer.

Don't use cos2x identities!

Look in your formula booklet for a formula for converting sinAcosB into the sum of 2 other trig functions, then integrate this

Checked answer and it was simplified down to

sin3xcos2x

1/2(sin5x + sinx)

= 1/2(-1/5cos5x - cosx) + c

= -1/10(cos5x +5cosx) + c


I used all the Cos2x identities and it still won't simplify to that answer.

Don't use cos2x identities!

Look in your formula booklet for a formula for converting sinAcosB into the sum of 2 other trig functions, then integrate this

Also how would you integrate (x+1)e^(x^2+2x) ?

That's e to the power of x squared + 2x (in case I wrote it wrong).

Answer is 1/2.e^(x^2+2x)

this is no longer in any formula booklets
you have to derive it, which is very easy

I'd be very surprised if it isn't somewhere in OCR C1-C4 - it's a very standard technique! Check your chapters on trig identities



you should be able to do this almost by recognition! What would you get if you differentiated e^(x^2 + 2x)? Can you see why this helps?

They're in the Edexcel and AQA formula booklets.

Also how would you integrate (x+1)e^(x^2+2x) ?

That's e to the power of x squared + 2x (in case I wrote it wrong).

Answer is 1/2.e^(x^2+2x)

Integration by parts

After I differentiate it I get (2x +2) then when I times that by (x +1) I get 2x^2+2x+2. What would you do from there?

Integration by parts

Thanks! The ans. for some reason did not have it.

That's because the "answer" is using the technique I explained above i.e, formula for converting product of 2 trig functions into a sum