Integration Techniques for the Function sin(x + π/6)cosx

[Nyasha]

Homework Statement

$$\int sin(x+\frac{\pi}{6})cosxdx$$

The Attempt at a Solution

$$u= sin(x+\frac{\pi}{6})$$

$$du=cosxdx$$

$$\frac{du}{cos x}=dx$$

$$\int udu=\frac{1}{2}sin(x+\frac{\pi}{6})+c$$

Guys how come my answer is different from the one at the back of the book. I'm l the one wrong or this time it is the book ?

 

[Dick]

You are, this time. If u=sin(x+pi/6) then du=cos(x+pi/6)*dx. The simple substitution doesn't work. Try using sin(a+b)=sin(a)*cos(b)+cos(a)*sin(b) on sin(x+pi/6) before integrating.

 

[lurflurf]

sin(x+pi/6)cos(x)=(1/2)(sin(2x+pi/6)+sin(pi/6))

 

[Nyasha]

You are, this time. If u=sin(x+pi/6) then du=cos(x+pi/6)*dx. The simple substitution doesn't work. Try using sin(a+b)=sin(a)*cos(b)+cos(a)*sin(b) on sin(x+pi/6) before integrating.

$$\int (sinx cos\frac{\pi}{6}+sin\frac{\pi}{6} cos x)cosx$$

$$\int sinx cosx cos\frac{\pi}{6}+\int sin \frac {\pi}{6} cos^2x$$

$$\frac{1}{2}cos\frac{\pi}{6}\int sin(2x)+\frac{1}{2}sin\frac{\pi}{6}\int(1+cos2x)$$

Is this now correct ?

 

[Dick]

$$\int (sinx cos\frac{\pi}{6}+sin\frac{\pi}{6} cos x)cosx$$

$$\int sinx cosx cos\frac{\pi}{6}+\int sin \frac {\pi}{6} cos^2x$$

$$\frac{1}{2}cos\frac{\pi}{6}\int sin(2x)+\frac{1}{2}sin\frac{\pi}{6}\int(1+cos2x)$$

Is this now correct ?

Yes, that looks right.

 

[Nyasha]

Yes, that looks right.

Thanks a lot for the help