Question Number 226995 by Spillover last updated on 24/Dec/25
Answered by Ghisom_ last updated on 24/Dec/25
![∫_0 ^(π/2) ((sin^3 x cos^5 x)/((1−(1/2)sin^2 x)^3 ))dx=∫_0 ^(π/2) (((1+cos 2x)sin^3 2x)/((3+cos 2x)^3 ))dx= [t=3+cos 2x → dx=−(dt/(2sin 2x))] =2∫_4 ^2 (((t−4)(t−2)^2 )/t^3 )dt=2∫_2 ^4 ((8/t)−((20)/t^2 )+((16)/t^3 )−1)dt= =2[8ln t +((20)/t)−(8/t^2 )−t]_2 ^4 =−11+16ln 2](https://www.tinkutara.com/question/Q226996.png)
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{sin}^{\mathrm{3}} \:{x}\:\mathrm{cos}^{\mathrm{5}} \:{x}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:{x}\right)^{\mathrm{3}} }{dx}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right)\mathrm{sin}^{\mathrm{3}} \:\mathrm{2}{x}}{\left(\mathrm{3}+\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{3}+\mathrm{cos}\:\mathrm{2}{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{2sin}\:\mathrm{2}{x}}\right] \\ $$$$=\mathrm{2}\underset{\mathrm{4}} {\overset{\mathrm{2}} {\int}}\frac{\left({t}−\mathrm{4}\right)\left({t}−\mathrm{2}\right)^{\mathrm{2}} }{{t}^{\mathrm{3}} }{dt}=\mathrm{2}\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\left(\frac{\mathrm{8}}{{t}}−\frac{\mathrm{20}}{{t}^{\mathrm{2}} }+\frac{\mathrm{16}}{{t}^{\mathrm{3}} }−\mathrm{1}\right){dt}= \\ $$$$=\mathrm{2}\left[\mathrm{8ln}\:{t}\:+\frac{\mathrm{20}}{{t}}−\frac{\mathrm{8}}{{t}^{\mathrm{2}} }−{t}\right]_{\mathrm{2}} ^{\mathrm{4}} =−\mathrm{11}+\mathrm{16ln}\:\mathrm{2} \\ $$
Commented by klipto last updated on 24/Dec/25
$$\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{bro}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{guys}}\:\boldsymbol{\mathrm{drop}}\:\boldsymbol{\mathrm{some}}\:\boldsymbol{\mathrm{other}} \\ $$$$\boldsymbol{\mathrm{textbooks}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{calculus}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{utilize}}? \\ $$
Answered by Spillover last updated on 24/Dec/25
Answered by Spillover last updated on 24/Dec/25
