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Verify by differentiation that: \[\LARGE \int (cos^3x)dx=sinx-\frac{1}{3}sin^3x+C\]
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Simply differentiate \[\frac{ 1 }{ 3 } \sin^{3}x\]
I never did understand how to differentiate with trig functions that have exponents. Could you walk me through it? .-.
use the chain rule. For d/dx f(u) take f'(u)u'
you will have to play with the trigonometric identities (I see a bit of sin^2 x+cos^2 x=1 in this one) to get what you want.
I leave it to inkyvod
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Why bother?
The silhouette man is here.
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