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What is the integral of #sin^5 (x) * cos^3 (x)#?

Answer 1

Paisley Johnson

#int sin^5 x*cos ^3 x d x=1/6sin^6 x-1/8 sin^8 x+C#

#int sin^5 x*cos ^3 x d x=?#

#int sin^5 x* cos^2x *cos x d x#

#cos^2 x=1-sin^2 x#

#int sin^5 x(1-sin^2 x) cos x d x#

#int (sin^5x-sin^7x)cos x d x#

#sin x=u" ; " cos x d x=d u#

#int (u^5-u^7)d u#

#int sin^5 x*cos ^3 x d x=1/6 u^6-1/8 u^8 +C#

#int sin^5 x*cos ^3 x d x=1/6sin^6 x-1/8 sin^8 x+C#

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Answer 2

Emilia Aguilar

To find the integral of sin^5(x) * cos^3(x), you can use the substitution method. Let u = sin(x), then du = cos(x) dx. Rewrite the integral in terms of u:

∫sin^5(x) * cos^3(x) dx = ∫u^5 * (1 - u^2) du.

Now, expand and integrate:

= ∫(u^5 - u^7) du = (1/6)u^6 - (1/8)u^8 + C.

Replace u with sin(x) to get the final answer:

= (1/6)sin^6(x) - (1/8)sin^8(x) + C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.