# Thread: What is wrong with my solution?

1. ## What is wrong with my solution?

Given:

$\int sin \frac{x}{2}cos \frac{x}{2}$

$-2 \int udu$

$-2\frac{u^{2}}{2}+c$

$-u^{2}+c$

$-cos\frac{x}{2}^{2}+c$

$-\frac{1}{2}cos[x]$

What's wrong with my solution?

any help will be appreciated, thanks

2. your solution is actually correct with the method you applied

to get the other answer make use of the identity $\sin 2x=2\sin x\cos x$ to subsitute and then integrate

3. Originally Posted by honestliar
Given:

$\int sin \frac{x}{2}cos \frac{x}{2}$

$-2 \int udu$

$-2\frac{u^{2}}{2}+c$

$-u^{2}+c$

$-cos\frac{x}{2}^{2}+c$

$-\frac{1}{2}cos[x]$

What's wrong with my solution?

any help will be appreciated, thanks
The simplest approach is to note that the integrand is the same as $\frac{1}{2} \sin (x)$ so you get $-\frac{1}{2} \cos (x) + K$ as the answer.

Now you should note that the difference between this and your answer of $- \cos \left( \frac{x}{2}\right) ^{2}+c$ (note the brackets by the way) is merely a constant ....

4. Originally Posted by mr fantastic
The simplest approach is to note that the integrand is the same as $\frac{1}{2} \sin (x)$ so you get $-\frac{1}{2} \cos (x) + K$ as the answer.

Now you should note that the difference between this and your answer of $- \cos \left( \frac{x}{2}\right) ^{2}+c$ (note the brackets by the way) is merely a constant ....

But the given is not just $\frac{1}{2} \sin (x)$

5. Recall the identity:

$sin(x) = 2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\r ight)$

6. Thank you, I got it.