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Question Number 84246 by pew_247 last updated on 10/Mar/20

∫cos^n  mx  dx =

$$\int\mathrm{cos}^{{n}} \:{mx}\:\:{dx}\:= \\ $$

Answered by TANMAY PANACEA last updated on 10/Mar/20

I_(n,m) =∫cos^n mxdx=∫cos^(n−1) mx.cosmxdx  cos^(n−1) mx.((sinmx)/m)+∫(n−1)cos^(n−2) mx.msinmx.((sinmx)/m)dx  ((cos^(n−1) mxsinmx)/m)+(n−1)∫cos^(n−2) mx(1−cos^2 mx)  I_(n,m) =((cos^(n−1) mx.sinmx)/m)+(n−1)I_(n−2,m) −(n−1)I_(n,m)    I_(n,m) =((cos^(n−1) mx.sinmx)/(mn))+((n−1)/n)I_(n−2,m)   pls check

$${I}_{{n},{m}} =\int{cos}^{{n}} {mxdx}=\int{cos}^{{n}−\mathrm{1}} {mx}.{cosmxdx} \\ $$$${cos}^{{n}−\mathrm{1}} {mx}.\frac{{sinmx}}{{m}}+\int\left({n}−\mathrm{1}\right){cos}^{{n}−\mathrm{2}} {mx}.{msinmx}.\frac{{sinmx}}{{m}}{dx} \\ $$$$\frac{{cos}^{{n}−\mathrm{1}} {mxsinmx}}{{m}}+\left({n}−\mathrm{1}\right)\int{cos}^{{n}−\mathrm{2}} {mx}\left(\mathrm{1}−{cos}^{\mathrm{2}} {mx}\right) \\ $$$${I}_{{n},{m}} =\frac{{cos}^{{n}−\mathrm{1}} {mx}.{sinmx}}{{m}}+\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2},{m}} −\left({n}−\mathrm{1}\right){I}_{{n},{m}} \: \\ $$$${I}_{{n},{m}} =\frac{{cos}^{{n}−\mathrm{1}} {mx}.{sinmx}}{{mn}}+\frac{{n}−\mathrm{1}}{{n}}{I}_{{n}−\mathrm{2},{m}} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}} \\ $$$$ \\ $$

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