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

William Collins 2022-01-03 Answered
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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Expert Answer

usumbiix
Answered 2022-01-04 Author has 1781 answers
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."
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usumbiix
Answered 2022-01-05 Author has 1781 answers
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.
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Vasquez
Answered 2022-01-08 Author has 9499 answers

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

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