∫ sin³ x cos³ x dx
∫sin3xcos3xdx\int \sin^3 x \cos ^3 x dx∫sin3xcos3xdx
Let t=cosxt =\cos xt=cosx , dt=−sinxdxdt = - \sin x dxdt=−sinxdx . Then ∫sin3xcos3xdx=∫−sin2x t3dt\int \sin^3 x \cos ^3 x dx = \int -\sin^2x \, t^3 dt∫sin3xcos3xdx=∫−sin2xt3dt
Also, sin2x=1−cos2x=1−t2\sin^2x = 1 -\cos^2x = 1-t^2sin2x=1−cos2x=1−t2 . So integral can be re-written as
∫−(1−t2)t3dt=−∫(t3−t5)dt=−t44+t66+C=cos6x6−cos4x4+C\displaystyle \int -(1-t^2) t^3 dt = - \int(t^3-t^5)dt = \frac{-t^4}{4} + \frac{t^6}{6} + C = \frac{\cos^6x}{6} - \frac{\cos^4x}{4}+C∫−(1−t2)t3dt=−∫(t3−t5)dt=4−t4+6t6+C=6cos6x−4cos4x+C
Answer: cos6x6−cos4x4+C\displaystyle \frac{\cos^6x}{6} - \frac{\cos^4x}{4}+C6cos6x−4cos4x+C
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