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Intergration (1 Viewer)

Thread starter Jimmy Marshman
Start date Jun 22, 2009

J

Jimmy Marshman

New Member

#1

Hey how do u answer this question.. i have tried everything

Find the integral of sin<SUP>2</SUP>xcos<SUP>2</SUP>x
any help appreciated

jet

Banned

#2

$\frac{1}{2}sin \, 2x = sin \, x cos \, x \\ \therefore \int sin^2 \,x cos^2 \, x \, dx = \frac{1}{4}\int sin^2 \, 2x \, dx \\ = \frac{1}{4} \left [\frac{1}{2}x - \frac{1}{8}sin \, 4x \right ] + C \text{ using the formula for the integral of } sin^2 \, ax$

Last edited: Jun 22, 2009

D

Drongoski

Well-Known Member

#3

$\80dpi \int sin^{2}x\ cos^{2}x\ dx\ \ =\ \ \int (\frac{1}{2}sin2x)^{2}\ dx\ \ =\ \ \frac{1}{8}\int (1-cos4x)\ dx\\ \\ \ =\ \ \frac{x}{8}\ -\ \frac{1}{32}sin4x\ +\ C$

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