Integrate sinx*cox* using double angle

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Homework Statement



∫ Sinx cos x dx

Homework Equations


The Attempt at a Solution


If you integrate it using substitution, you get -cos2(x)/2but if you use double angle formula to rewrite the problem, it will be ∫1/2 sin(2x), and integrate it, you get -cos(2x)/4. isn't it weird?
 
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You are forgetting that each of those two integrals must have ##+ C## appended to it. Now recall that ## \cos 2x = 2 \cos^2 x - 1 ##.
 
Jude075 said:
If you integrate it using substitution, you get -cos2(x)/2but if you use double angle formula to rewrite the problem, it will be ∫1/2 sin(2x), and integrate it, you get -cos(2x)/4. isn't it weird?
What voko said is exactly right. To elaborate, these two apparently different integrals differ only by a constant.
 
So if I was to find the indefinite integral, both answers were right?
 
None is correct unless you add the indefinite constant.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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