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ammu123
evaluate the integral (integral sign) cos x dx/ 1+8 sinx
Use U-Substitution. Let u = 1+8sinx.
can u explain how to do it please.
let t=1+8sin(x) then dt=8cos(x)dx 1/8dt=cos(x)dx integral becomes \[\Large \frac{1}{8}\int\limits_{}^{} \frac{1}{t}dt\] you may know that formula that \[\Large \int\limits_{}^{} \frac{1}{x}dx =\ln(x)+C\] use this to evaluate hte above integral then substitute back t=1+8sin(x)