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help me with this further integration q pls (1 Viewer)

Thread starter maxxxxx
Start date May 28, 2024

M

maxxxxx

New Member

#1

like nothing ive ever seen before, or i might be completely trippin, please help

Use the substitution u = cos x to find ∫sin^3 x dx

T

tywebb

dangerman

#2

$du=-\sin x\ \!dx$

$\therefore\textstyle\int\sin^3x\ \!dx=\int-\sin^2x(-\sin x\ \!dx)=\int(\cos^2x-1)(-\sin x\ \!dx)$

$\textstyle=\int(u^2-1)\ \!du=\frac{u^3}{3}-u+c=\frac{\cos^3x}{3}-\cos x+c$

K

kareem!

New Member

#3

the odd one out

T

tywebb

dangerman

#4

kareem! said:
the odd one out

so were you the first time you saw it

M

maxxxxx

New Member

#5

tywebb said:
$du=-\sin x\ \!dx$

$\therefore\textstyle\int\sin^3x\ \!dx=\int-\sin^2x(-\sin x\ \!dx)=\int(\cos^2x-1)(-\sin x\ \!dx)$

$\textstyle=\int(u^2-1)\ \!du=\frac{u^3}{3}-u+c=\frac{\cos^3x}{3}-\cos x+c$

you are the best, tysm

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