Integral Calculus: Integration by Substitution. Final answers should be in its simplest form.
Use the solved problem as your reference.
Transcribed Image Text:X = Za sina
dx
8. S
xV2ax-x2
dx =2.2 a sina cosodə
at -xZ
dx = 4a sina cos ode
2
X = la sina
4a sine cos e do
Zax-x2
Xo Za tato rae to?
Lasints)
za (tasw'o) - (2osinto)²
X -la seco -lax
Vx²-Zax
4a sin o coso do
Za sin?o
Vta? sinto - 4a sintə
tasıno cosede
Za sinto
Ha sinto Li-sıno)
4a sino coso do
Za sinto
a sina
Cosa do
X = la sinta
sino
Vzax
za
ax
Sin e =
Vea
a
Sino coso
Ir do
sinta
-2
sino de
a
I Sin a
arx
do
Sin e
%3D
La-x
a
cscode
la-x
coto tC
arx
C=
X - Za zinto
lax -x2
ax
24
tano=
PINAL ANSWER
za
coto
2a - X
Transcribed Image Text:dx
(Use problem 8 from
as your reference)
xVx2-2ax
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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