Question

Integrate [sinx]^3 then show that 1/3[cosx]^3-1/4cosx-1/12cos3x+k=0

Answer 1

i have done the integration part how could i prove that the integration of [sinx]^3 is equivalent 1/3[cosx]^3-1/4cosx-1/12cos3x+k

Answer 2

\[\int\limits_{}^{} (sinx)^3=\int\limits_{}^{}sinx*(sinx)^2=\int\limits_{}^{}sinx(1-(cosx)^2)=\int\limits_{}^{}=sinx-(cosx)^2sinx\]

Answer 3

let u=cosx

Answer 4

\[du/dx=-sinx \rightarrow dx=-du/sinx\]
\[\int\limits_{}^{}sinx-u^2sinx*-du/sinx=sinx+u^2\]

Answer 5

\[\int\limits_{}^{}sinx+u^2=-cosx+u^3/3=-cosx+(cosx)^3/3\]

Answer 6

use this.. sin 3x = 3 sin x - 4sin^3 x

Answer 7

3sinx-sin3x)1/4=sin^3x

Answer 8

integrate 1/4(3sinx-sin3x)
3/4sinx-1/4sin3x
-3/4cosx+1/12cos3x
1/4(-3cosx+1/3cos3x)

Answer 9

seems good..

Answer 10

@ganeshie8

Answer 11

@Abmon98 what you've done is correct. if the original integral be A the second method has given you A. Just subtract both the equations you get
A - A = (1/3[cosx]^3 - cos x + K') - (-3/4[cos x] +1/12 [Cos3x] + K" )

Answer 12

K' and K" are the constants of Indefinite integration. Their difference is another constant K.

Answer 13

C(constant of first method)-C'(constant of second method)=K
is this right