+0

# trigonometric integration

0
198
2
+1828

which one is correct and why

xvxvxv  Feb 10, 2015

#1
+80875
+10

d/dx ( sin^4/4 - sin^6/6 + C) =

sin^3x(cosx) - sin^5x(cosx) =

cosx(sin^3x)(1 - sin^2x) =

cosx (sin^3x)(cos^2x) =

sin^3xcos^3x

And

d/dx(-cos^4x/4 + cos^6x/6 + C)=

-(sinx)(-cos^3x)   - (sinx)cos^5x  =

(sinx)(cos^3x)(1 - cos^2x) =

(sinx)(cos^3x)(sin^2x) =

(cos^3x)(sin^3x)  = sin^3xcos^3x

And they are exactly the same.....!!!

CPhill  Feb 10, 2015
Sort:

#1
+80875
+10

d/dx ( sin^4/4 - sin^6/6 + C) =

sin^3x(cosx) - sin^5x(cosx) =

cosx(sin^3x)(1 - sin^2x) =

cosx (sin^3x)(cos^2x) =

sin^3xcos^3x

And

d/dx(-cos^4x/4 + cos^6x/6 + C)=

-(sinx)(-cos^3x)   - (sinx)cos^5x  =

(sinx)(cos^3x)(1 - cos^2x) =

(sinx)(cos^3x)(sin^2x) =

(cos^3x)(sin^3x)  = sin^3xcos^3x

And they are exactly the same.....!!!

CPhill  Feb 10, 2015
#2
+1828
0

Thank you CPhill

xvxvxv  Feb 13, 2015

### 25 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details