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




Question Number 13000 by Joel577 last updated on 10/May/17
∫ sin^4  x cos^3  x dx
$$\int\:\mathrm{sin}^{\mathrm{4}} \:{x}\:\mathrm{cos}^{\mathrm{3}} \:{x}\:{dx} \\ $$
Commented by Joel577 last updated on 10/May/17
What is the idea to solve that problem?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{idea}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{that}\:\mathrm{problem}? \\ $$
Commented by prakash jain last updated on 10/May/17
sin^4 x cos^3 x  =sin^4 x(1−sin^2 x)cos x  put sin x=u  cos xdx=du
$$\mathrm{sin}^{\mathrm{4}} {x}\:\mathrm{cos}^{\mathrm{3}} {x} \\ $$$$=\mathrm{sin}^{\mathrm{4}} {x}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)\mathrm{cos}\:{x} \\ $$$${put}\:\mathrm{sin}\:{x}={u} \\ $$$$\mathrm{cos}\:{xdx}={du} \\ $$
Commented by Joel577 last updated on 11/May/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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