Please Help. A-maths (integration)

Ajimal SG

SOLVED !!!!

This Question is from my textbook. (Panpac, Addditional-Maths)

Misc. Examples 19, Q16(b), Pg.422.

Find d/dx (sin^3 2x) and hence evaluate the integral(∫) from (π/4) to (0). (cos^3 2x) dx.

Note: d/dx (sin^3 2x) = (3sin^2 2x)(2cos 2x) = 6(sin^2 2x)(cos 2x)

PLEASE HELP!!!

Ajimal SG

OOOPS!!! I forgot to tell u the answer XD...

The Answer is (1/3).

eagle SG

$\int_{0}^{\pi /4}{\cos^{3} 2x }dx$
$=\int_{0}^{\pi /4}{\cos 2x \left (\cos^{2} 2x \right )}dx$
$=\int_{0}^{\pi /4}{\cos 2x \left (1-\sin^{2} 2x \right )}dx$
$=\int_{0}^{\pi /4}{\cos 2x -\cos 2x\sin^{2} 2x }dx$
$=\int_{0}^{\pi /4}{\cos 2x }dx-\int_{0}^{\pi /4}{\cos 2x \sin^{2}2x}dx$

You should be able to continue from here

^tamago^ SG

OMG I’ve returned them all. Surprised

Ajimal SG

Yes eagle, that is true. But u will have to use d/dx (sin^3 2x) to solve the question. Thats why i bold the word "hence" in the question. Thanks for ur concern anyways :D...

Ajimal SG

Nvm all... i solved the question already :D... Thx for ur acknowledgement anyways...

Do u guys realise that i spelled Additional maths wrongly??? It suppose to have only 2 "d"s... But i spelled "Addditional" with 3 "d"s.... XD... juz for ur info...

eagle SG

you still have to use it in this portion

$\int_{0}^{\pi /4}{\cos 2x \sin^{2}2x}dx$

because integrating it will give you $\frac{1}{6}\sin^{3}2x$

because you differentiated at first to get $6\sin^{2}2x \cos 2x$

Ajimal SG

Yep yep

Well... this is how i solved the question...

6∫(sin^2 2x)(cos 2x) = sin^3 2x

6∫(cos 2x - cos^3 2x) = sin^3 2x

-6∫(-cos 2x + cos^3 2x) = sin^3 2x

∫-cos 2x + ∫cos^3 2x) = (1/-6)(sin^3 2x)

∫(cos^3 2x) = [(1/-6)(sin^3 2x)] / (cos^3 2x)

Then continue from here :D .... lazy type it out...

6∫(1-cos^2 2x)(cos 2x) = sin^3 2x