trigonometry and integration (1 Viewer)

[sasquatch]

A question asks: Find d/dx(x sin2x) and hence find ₀∫^π/4 x cos2x dx.

Heres my working out

Let f(x) = x sin 2x
f'(x0 = sin2x + 2xcos2x

₀∫^π/4 (x cos2x) dx
= 1/2 ₀∫^π/4 (2x cos 2x) dx
= 1/2 ₀∫^π/4 (f'(x) - sin2x) dx
= 1/2 [sin 2x + cos2x / 2]sub]0[/sub]^π/4
= 1/2 [1 + 0 - 0 - (1/2)
= 1/2 * 1/2
= 1/4

But that is not correct. I have no actual idea how to do this question, i just took a guess.

Could anybody lend a hand? Thanks.

WOOPS sorry, working out error. Ive got it now..

 

[pLuvia]

y = xsin2x
y'= 2xcos2x + sin2x

₀∫^π/4 x cos2x dx
= 1/2 ₀∫^π/4 2xcos2x dx
= 1/2 ₀∫^π/4 y' - sin2x dx
= 1/2 [ xsin2x - sin2x ] Sub π/4 and 0
= 1/2 [ (π/4 - 1) ]

 

[sasquatch]

WOOPS sorry, working out error. Ive got it now..

Thanks anyway though.

 

[Cardscook77]

Could you explain why you don't also integrate the -sin2x in the brackets which you sub into?

 

[B1andB2]

Could you explain why you don't also integrate the -sin2x in the brackets which you sub into?

you're right, they forgot to integrate in the second last line. It should be

1/2 [ xsin2x + cos2x/2 ] Sub π/4 and 0

 

[011235]

Lol 15 years later

 

[Cardscook77]

Yeah thought I was doing something wrong (good to clear that up). Thanks for the responses on a 15-year-old thread!