# I tried it using the double angle identity 2 \sin x \

I tried it using the double angle identity
$$\displaystyle{2}{\sin{{x}}}{\cos{{x}}}$$
The answer that I got is
$$\displaystyle{\frac{{-{\cos{{2}}}{x}}}{{{4}}}}+{c}$$
However I've also tried it using u-subsitution.
I let $$\displaystyle{u}={\sin{{x}}}$$. Thus obtaining $$\displaystyle{\cos{{x}}}$$ when differentiating. And cutting the $$\displaystyle{\cos{{x}}}\ \text{ in }\ {2}{\sin{{x}}}{\cos{{x}}}$$ out with the $$\displaystyle{\cos{{x}}}$$ in the denominator below du.
However the answer that I am then getting is : $$\displaystyle{0.25}-{0.25}{\cos{{2}}}{x}+{c}$$

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usumbiix
Both answers are correct. You can use double angle identity, as well as u sub for either $$\displaystyle{\sin{{x}}}\ \text{ or }\ {\cos{{x}}}$$
The key lies in the +c. All the 3 integrals are a family of functions just separated by a different "+c". In practice, double angle identity is often used as it's more intuitive and simpler in some sense. But the other methods are perfectly acceptable, and not "wrong."
###### Not exactly what you’re looking for?
usumbiix
Actually your method is not wrong.
All you need to do is just substitute
$$\displaystyle{c}_{{2}}={0.25}+{c}$$
and then you get the "correct" answer.
Vasquez

$$\frac12 \frac{d}{dx} \sin^2 x=\sin x \cos x$$
so
$$\int \sin x \cos x dx=\frac12 \sin^2 x+\text{constant}$$