differential trig problem

**slaypullingcat** · April 30th 2009, 11:27 PM

Show that d/dx (1/5 sin^5 x - 1/7 sin^7 x) = sin^4 x cos^3 x
Also, I typed out this question in word using equation editor and tried to paste it in without luck (just thought it would be easier for helpers?). I'm not sure how to paste in the actual question, or, is the above acceptable?
Thanks all the same.

**Spec** · April 30th 2009, 11:40 PM

Here's a good thread about the math code this forum uses: http://www.mathhelpforum.com/math-he...-tutorial.html

On to your problem:

$sin^4xcos^3x = sin^4x(1-sin^2x)cos x = cosx(sin^4x-sin^6x)$

The last part is easy to integrate since $\int cosxsin^{n-1}x dx = \frac{1}{n}sin^nx$

**redsoxfan325** · April 30th 2009, 11:44 PM

Originally Posted by slaypullingcat

Show that d/dx (1/5 sin^5 x - 1/7 sin^7 x) = sin^4 x cos^3 x
Also, I typed out this question in word using equation editor and tried to paste it in without luck (just thought it would be easier for helpers?). I'm not sure how to paste in the actual question, or, is the above acceptable?
Thanks all the same.

Show that: $\frac{d}{dx}\left[\frac{1}{5}\sin^5(x)-\frac{1}{7}\sin^7(x)\right] = \sin^4(x)\cos^3(x)$

$\frac{d}{dx}\left[\frac{1}{5}\sin^5(x)\right] = 5\cdot\frac{1}{5}\sin^4(x)\cdot\cos(x) = \sin^4(x)\cos(x)$

$\frac{d}{dx}\left[\frac{1}{7}\sin^7(x)\right] = 7\cdot\frac{1}{7}\sin^6(x)\cdot\cos(x) = \sin^6(x)\cos(x)$

So the derivative is $\sin^4(x)\cos(x)-\sin^6(x)\cos(x) = \sin^4(x)\cos(x)\cdot(1-\sin^2(x))$ $= \sin^4(x)\cos(x)\cdot\cos^2(x) = \sin^4(x)\cos^3(x)$

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