dx-sin-3-x-cos-5-x-

Question Number 93175 by i jagooll last updated on 11/May/20

$$\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{3}} \:\left(\mathrm{x}\right).\mathrm{cos}\:^{\mathrm{5}} \left(\mathrm{x}\right)}}\:?\: \\ $$

Commented by i jagooll last updated on 11/May/20

what is the idea to solve this problem, prof mr mjs?

Answered by M±th+et+s last updated on 11/May/20

∫(1/(sin^(3/2) (x)cos^(5/2) (x))). ((1/cos^(3/2) (x))/(1/cos^(3/2) (x)))dx ∫((sec^2 (x))/(tan^(3/2) (x)cos^2 (x)))dx ∫((tan^2 (x)+1)/(tan^(3/2) (x)cos^2 (x)))dx ∫((tan^2 (x))/(tan^(3/2) (x)cos^2 (x)))dx+∫(1/(tan^(3/2) (x)cos^2 (x)))dx ∫((√(tan(x)))/(cos^2 (x)))dx+∫tan^((−3)/2) sec^2 (x)dx =(2/3)tan^(3/2) (x)−2tan^((−1)/2) (x)+c

$$\int\frac{\mathrm{1}}{{sin}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right){cos}^{\frac{\mathrm{5}}{\mathrm{2}}} \left({x}\right)}.\:\frac{\mathrm{1}/{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right)}{\mathrm{1}/{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right)}{dx} \\ $$$$\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\int\frac{{tan}^{\mathrm{2}} \left({x}\right)+\mathrm{1}}{{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\int\frac{{tan}^{\mathrm{2}} \left({x}\right)}{{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)}{dx}+\int\frac{\mathrm{1}}{{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\int\frac{\sqrt{{tan}\left({x}\right)}}{{cos}^{\mathrm{2}} \left({x}\right)}{dx}+\int{tan}^{\frac{−\mathrm{3}}{\mathrm{2}}} {sec}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}\right)−\mathrm{2}{tan}^{\frac{−\mathrm{1}}{\mathrm{2}}} \left({x}\right)+{c} \\ $$

Commented by i jagooll last updated on 11/May/20

thank you. but it should be −2 tan^(−1/2) x sir

$$\mathrm{thank}\:\mathrm{you}.\:\mathrm{but}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be} \\ $$$$−\mathrm{2}\:\mathrm{tan}\:^{−\mathrm{1}/\mathrm{2}} {x}\:{sir}\: \\ $$

Commented by M±th+et+s last updated on 11/May/20

$${yes}\:{sir}\:{its}\:{a}\:{typo}\: \\ $$

Answered by john santu last updated on 11/May/20

Commented by i jagooll last updated on 11/May/20

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by niroj last updated on 11/May/20

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Answered by MJS last updated on 11/May/20

∫(dx/( (√(sin^3 x cos^5 x))))= [t=tan x → dx=cos^2 x dt] =∫((t^2 +1)/t^(3/2) )dt=∫t^(1/2) dt+∫t^(−3/2) dt= =(2/3)t^(3/2) −2t^(−1/2) =...

$$\int\frac{{dx}}{\:\sqrt{\mathrm{sin}^{\mathrm{3}} \:{x}\:\mathrm{cos}^{\mathrm{5}} \:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{3}/\mathrm{2}} }{dt}=\int{t}^{\mathrm{1}/\mathrm{2}} {dt}+\int{t}^{−\mathrm{3}/\mathrm{2}} {dt}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}/\mathrm{2}} −\mathrm{2}{t}^{−\mathrm{1}/\mathrm{2}} =… \\ $$

Commented by i jagooll last updated on 11/May/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$

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