Sin(2x) + 2 differentiates to 2cos(2x), so you can use the ln(f(x)) rule to integrate it to [ 0.5ln( sin(2x) + 2 ) ], and then the rest of the question is just simplifying until you get an answer of 0.5ln(5/4)

Reply 7

BigJ123

10

Original post by zac777

I think we are both wrong, i got ln(1/12 pie) and you got ln root5/2) but that doesnt match the calculator's integration function

No I’m correct, not sure what you did but I’m certainly correct ( it does match the calculator )

Reply 8

BigJ123

10

Original post by Throwaway1686

Sin(2x) + 2 differentiates to 2cos(2x), so you can use the ln(f(x)) rule to integrate it to [ 0.5ln( sin(2x) + 2 ) ], and then the rest of the question is just simplifying until you get an answer of 0.5ln(5/4)

Been a while since I’ve done any integration so I didn’t immediately spot that but very good

Reply 9

zac777

6

Original post by Throwaway1686

Sin(2x) + 2 differentiates to 2cos(2x), so you can use the ln(f(x)) rule to integrate it to [ 0.5ln( sin(2x) + 2 ) ], and then the rest of the question is just simplifying until you get an answer of 0.5ln(5/4)

I did u=sin2x
and got integral (1/2 to 0) with 1/(u+2) du
Why is this wrong?

Reply 10

BigJ123

10

Original post by zac777

I did u=sin2x
and got integral (1/2 to 0) with 1/(u+2) du
Why is this wrong?

du/dx = 2cos2x ==> 1/2 2cos2x du = dx. Your integral then should be 1/2 x The integral from (0 to 1/2) of 1/(u+2) du

Your limits are the wrong way around also.

Hard integration Q

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