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Help with AQA core 4 integration (AQA spec)

Basically I have found some integration practice questions on the internet but they have conveniently failed to provide the answers. I am confident on most of them that I have go the correct answer however there are a few which I'm unsure on. So I was hoping the lovely people of the student room could help out, I've provided all my working too, so could you please tell me where I've gone wrong in these questions.

Thanks :smile:

q1.jpg

q2.jpg

q3.jpg

q4.jpg

Note: I've circled the questions in red because they can be well hidden and underlined my answers in red.

Please help XD

Thanks :smile:
Reply 1
Original post by Zozzy
Basically I have found some integration practice questions on the internet but they have conveniently failed to provide the answers. I am confident on most of them that I have go the correct answer however there are a few which I'm unsure on. So I was hoping the lovely people of the student room could help out, I've provided all my working too, so could you please tell me where I've gone wrong in these questions.

Thanks :smile:

q1.jpg

q2.jpg

q3.jpg

q4.jpg

Note: I've circled the questions in red because they can be well hidden and underlined my answers in red.

Please help XD

Thanks :smile:


About to go offline now, but for the very first one, note that

eudu=eu+C\int e^u du = e^u + C - you seem to have introduced an extra factor of 1/u in your answer!
Reply 2
Original post by davros
About to go offline now, but for the very first one, note that

eudu=eu+C\int e^u du = e^u + C - you seem to have introduced an extra factor of 1/u in your answer!


Ah thanks I can see the silly mistake now thanks :smile:
Original post by Zozzy

Spoiler



2 and 4 are fine, but 3 has some problems. You've made a bunch of arithmetical mistakes like multiplying things by negative 1 that don't need to be multiplied by negative 1.

It might be a bit easier to completely transform your integral from xx to uu as opposed to leaving in parts of xx and uu in the same integral.

sin32x dx=(1cos22x)sin2x dx=12(1cos22x)= u2(2)sin2x dx= du=u=cos2x 12(1u2) du\displaystyle \begin{aligned} \int \sin^3 2x \text{ d}x & = \int \left( 1 - \cos^2 2x \right) \sin 2x \text{ d}x \\ & = \dfrac{-1}{2} \int ( 1 - \underbrace{ \cos^2 2x )}_{= \text{ u}^2} \underbrace{ (-2) \sin 2x \text{ d}x }_{ = \text{ d}u} \\ & \overset{u = \cos 2x}= \ \dfrac{-1}{2} \int \left( 1 - u^2 \right) \text{ d}u \end{aligned}
(edited 9 years ago)
First can also be done by inspection btw. Notice that ddxe2x+2=(2x+2)e2x+2=2(x+1)e2x+2\frac{d}{dx}e^{2x+2} = (2x+2)e^{2x+2} = 2(x+1)e^{2x+2}.

Second is ok.

Third one I see you went from sin32x\sin^32x to sin22xsinx\sin^22x\sin x.

Last one is ok but maybe leave as 8log48\log4.
(edited 9 years ago)

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