Question

Integrate sin3xcos3x dx

Answer 1

will give a medal please need help

Answer 2

\[\int\limits \sin3xcos3x= \int\limits \frac{1}{2} \times 2\sin3xcos3x= \frac{1}{2} \int\limits \sin6x\]
Let 6x=t
Therefore:
6dx=dt
i.e. dx=1/6dt
Hence \[ \frac{1}{2} \int\limits \sin6x = \frac{1}{2} \int\limits sint \frac{dt}{6} = \frac{1}{2} \times \int\limits sint \frac{dt}{6} \]
\[= \frac{1}{2} \times \frac{1}{6} (-cost)+c = -\frac{1}{12}cost+c\]
\[= -\frac{1}{12}\cos6x+c\] (Since t=6x)

@07supatel

Answer 3

why is there 1/2 there

Answer 4

thats the bit i dont understand

Answer 5

@07supatel
We know that:
\[\sin2 \theta = 2\sin \theta \cos \theta \rightarrow \sin2 \theta = 2\sin \theta \cos \theta\]
From the formula it is clear that: In order to write sinθ cosθ as sin2θ we should have written 2sinθcosθ =sin2θ

i.e. 2sinθcosθ =sin2θ or vice versa
similarly \[\sin3xcos3x = \frac{1}{2} \times 2 \sin3xcos3x = \frac{1}{2} \times \sin(2 \times3x)\]
\[= \frac{1}{2} \sin(6x)\]

Here 2 is multiplied and divided to balance the ratio.