sin^2x and cos^2x (1 Viewer)

Ryuu-jin-jakka · Mar 23, 2009

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lychnobity · Mar 23, 2009

a) Convert cos<sup>2</sup>x into (cos 2x +1)/2

(3sin 2x)/2 + 3x/2 - 5x + c

b) 7sin2x/2 + tan7x +c

Ryuu-jin-jakka · Mar 23, 2009

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a) 3/4 sin2x – 7x/2 +C
<o

> b) </o

>7x/2 +7/4sin2x + 5x/2 +C

GUSSSSSSSSSSSSS · Mar 23, 2009

yea his method is right he just screwed up the working

forgot about f'(x) when integrating!

jet · Mar 23, 2009

$\int 3cos^2x - 5 \, dx \\ = 3(\frac{x}{2} + \frac{1}{4}sin2x) - 5x + C \\ = \frac{3}{4}sin2x + \frac{3x-10x}{2} + C \\ = \frac{3}{4}sin2x - \frac{7x}{2} + C \\ \\ \int 7cos^2x + sec^27x \, dx \\ = 7(\frac{x}{2} + \frac{1}{4}sin2x) + \frac{1}{7}tan7x + c \\ = \frac{7x}{2} + \frac{7}{4}sin2x + \frac{1}{7}tan^27x + C$
Dont know how you got that second answer, since $\int sec^2x \, dx = tanx + C$

Drongoski · Mar 23, 2009

Ryuu-jin-jakka · Mar 23, 2009

thanks you! poeople!!

sin^2x and cos^2x (1 Viewer)

Ryuu-jin-jakka

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