Question

Integration

Answer 1

\[\frac{ 18 }{ 2\sqrt{2}-2 }\int\limits_{0}^{\pi/12}[\cos^2(3x)-\sin^2(3x)]dx\]

How would I integrate this?

Answer 2

Half-Angle Formula's, ya? :)

Answer 3

\[\large\rm \cos^2(\color{orangered}{t})=\frac12[1+\cos(2\color{orangered}{t})]\]\[\large\rm \sin^2(\color{orangered}{t})=\frac12[1-\cos(2\color{orangered}{t})]\]

Answer 4

\[\frac{ 18 }{ 2\sqrt{2}-2 }\int\limits_{0}^{\pi/12}(\frac{ 1 }{ 2 })1+\cos3x)-\frac{ 1 }{ 2 }(1-\cos3x))dx\]\[\frac{ 9 }{ 2\sqrt{2}-2 }\int\limits_{0}^{\pi/12}(1+\cos3x-1+\cos3x)dx\]

Like that?

Answer 5

Woops, you didn't double the angle when you applied the formula :O
\[\large\rm \cos^2(\color{orangered}{3x})=\frac12[1+\cos(2\color{orangered}{\cdot3x})]\]

Answer 6

\[\frac{ 18 }{ 2\sqrt{2}-2 }\int\limits_{}^{}[\frac{ 1 }{ 2 }(1+\cos6x)-\frac{ 1 }{ 2 }(1-\cos6x)]dx\]

Like that?

Answer 7

Mmmm ya looks better!
Let's factor the 1/2 out to the front, and distribute the negative to the second set of brackets,\[\large\rm \frac12\cdot\frac{18}{2\sqrt{2}-2}\int\limits1+\cos6x-1+\cos6x~dx\]

Answer 8

cos^2 x - sin^2 x is cos(2x)
in this case, cos(6x)

Answer 9

Yes, applying our Double Angle Formula would have been easier now that I see it :D hehe

Answer 10

Okay, thanks zepdrix