1.

`int_0^((pi)/2) cos^5xsin^4x` dx

Answer» `I = int_(pi/2)^0 cos^5x sin^4x dx`
`=>I = int_(pi/2)^0 cos^4x sin^4xcosx dx`
`=>I = int_(pi/2)^0 (1-sin^2x)^2 sin^4xcosx dx`
Let `sinx = t`, then, `cosxdx = dt`
Then, our integral becomes,
`=> I = int_0^1 (1-t^2)^2t^4dt`
`=> I = int_0^1 (1+t^4-2t^2)t^4dt`
`=> I = int_0^1 (t^4+t^8-2t^6)dt`
`=>I = [t^5/5+t^9/9-(2t^7)/7]_0^1`
`=>I = (1/5+1/9-2/7)-(0+0-0) = (63+35-90)/315`
`=> I = 8/315`


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