need help (past paper question) (1 Viewer)

jason2kool · Mar 27, 2012

http://static.movember.com/uploads/...716b7802a7d061323ac9f8ae342-4f716d5548d0e.jpg

first question on the sheet.. sorry forgot to flip it when uploading.
thanks in advance

nightweaver066 · Mar 27, 2012

$We know that $ sin2x = 2sinxcosx$

$sin^2xcos^2x = sinxcosx \times sinxcosx$

$= \frac{1}{2}sin2x \times \frac{1}{2}sin2x$

$= \frac{1}{4}sin^2 2x$

$\therefore \int sin^2xcos^2x dx$

$= \frac{1}{4} \int sin^22xdx$

$= \frac{1}{8} \int (1 - cos4x) dx$

$And go from there$

SpiralFlex · Mar 27, 2012

$\frac{1}{4}\int_{0}^{\frac{\pi}{4}}(\sin2x)^2 \; dx$

$\frac{1}{8}\int_{0}^{\frac{\pi}{4}}(1-\cos 4x)\; dx$

$\frac{1}{8}[x-\frac{1}{4}\sin 4x]$

$\frac{\pi}{12}$

SpiralFlex · Mar 27, 2012

Nightweaver, You forgot to double the double angle

nightweaver066 · Mar 27, 2012

SpiralFlex said:
Nightweaver, You forgot to double the double angle

Yeah i've realised.. Been sick and making more mistakes lately :/

SpiralFlex · Mar 27, 2012

^Get well soon.

SpiralFlex · Mar 27, 2012

nightweaver066 · Mar 27, 2012

Thanks.

Was that bear on your pen a few weeks ago?

SpiralFlex · Mar 27, 2012

Correct.

imsleepyz · Mar 28, 2012

Hey SpiralFlex, the answer is pi/32 not pi/12 because pi/4 will multiply with 1/8

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need help (past paper question) (1 Viewer)

jason2kool

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