How do you find #\int \cos ^ { 5} x \sin ^ { 4} x d t#?

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Answer 1 · 2017-06-29T16:44:20

# sin^5x/315(35sin^4x-90sin^2x+63)+C.#

Let, #I=intcos^5xsin^4xdx.#

We subst., #sinx=t rArr cosxdx=dt.#

# :. I=intcos^5xsin^4xdx,#

#=intcos^4xsin^4xcosxdx,#

#=int(1-sin^2x)^2sin^4xcosxdx,#

#=int(1-t^2)^2t^4dt,#

#=int(1-2t^2+t^4)t^4dt,#

#=int(t^4-2t^6+t^8)dt,#

#=t^5/5-2*t^7/7+t^9/9,#

#=t^5/(5*7*9)(63-90t^2+35t^4),#

# rArr I=sin^5x/315(35sin^4x-90sin^2x+63)+C.#

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