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cos-4-x-sin-3-x-dx-




Question Number 7958 by tawakalitu last updated on 25/Sep/16
∫cos^4 x sin^3 x  dx
$$\int{cos}^{\mathrm{4}} {x}\:{sin}^{\mathrm{3}} {x}\:\:{dx}\: \\ $$
Commented by sandy_suhendra last updated on 25/Sep/16
=∫cos^4 x sinx sin^2 x dx  =∫cos^4 x sinx (1−cos^2 x) dx  =∫cos^4 x sinx dx − ∫cos^6 x sinx dx  =− (1/5) cos^5 x + (1/7) cos^7 x + c
$$=\int{cos}^{\mathrm{4}} {x}\:{sinx}\:{sin}^{\mathrm{2}} {x}\:{dx} \\ $$$$=\int{cos}^{\mathrm{4}} {x}\:{sinx}\:\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)\:{dx} \\ $$$$=\int{cos}^{\mathrm{4}} {x}\:{sinx}\:{dx}\:−\:\int{cos}^{\mathrm{6}} {x}\:{sinx}\:{dx} \\ $$$$=−\:\frac{\mathrm{1}}{\mathrm{5}}\:{cos}^{\mathrm{5}} {x}\:+\:\frac{\mathrm{1}}{\mathrm{7}}\:{cos}^{\mathrm{7}} {x}\:+\:{c} \\ $$
Commented by tawakalitu last updated on 25/Sep/16
Thanks so much sir.
$${Thanks}\:{so}\:{much}\:{sir}. \\ $$
Answered by prakash jain last updated on 02/Oct/16
answer is comments
$$\mathrm{answer}\:\mathrm{is}\:\mathrm{comments} \\ $$

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