Question Number 11436 by @ANTARES_VY last updated on 26/Mar/17

  ∫_0 ^(𝛑/4) sinx×cos^7 x dx.  solves...

Commented byFilupS last updated on 26/Mar/17

let u=cos(x)  ∴ du=−sin(x)dx     ∴∫_0 ^( π/4) sin(x)cos^7 (x)dx=−∫_(x=0) ^( x=π/4) u^7 du  −∫_(x=0) ^( x=π/4) u^7 du=−[(1/8)u^8 ]_(x=0) ^(x=π/4)   =−(1/8)[cos^8 (x)]_(x=0) ^(x=π/4)   =−(1/8)(cos^8 ((π/4))−cos^8 (0))  =−(1/8)(((1/(√2)))^8 −1)  =−(1/8)(2^(−(1/2)×8) −1)  =−(1/8)(2^(−4) −1)  =−(1/2^3 )((1/2^4 )−1)  =−((1/2^7 )−(1/2^3 ))  =−((1/2^7 )−(2^4 /2^7 ))  =−(((1−2^4 )/2^7 ))  =((2^4 −1)/2^7 )  =((15)/(128))