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I have just begun my 2nd calculus course and so far have just been applying the substitution method for solving anti derivatives and other basic rules.

I have a question that is probably very easy to answer.

Is $\displaystyle\int \left(\sin^2x + \cos^2x\right)\;dx = \int 1 \; dx$ ?

Thanks, Sam

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3  
Pythagorean Theorem... (yes, just with algebra: $\sin^2 x+\cos^2 x=1$). –  David Mitra Feb 8 '12 at 22:39
4  
The trigonometric identities that you probably learned in previous years still apply to your calculus course... –  Guess who it is. Feb 8 '12 at 22:41
2  
But you'll probably make your teacher happier if you put in some parentheses: $\int(\sin^2x+\cos^2x)\,dx=\int1\,dx$. –  Gerry Myerson Feb 8 '12 at 23:06

2 Answers 2

Yes, that s right.

You don't even need to use any of the calculus you may have seen in the course, it just follows straight from the fact, that $\sin^2x + \cos^2x = 1 $

To see this just consider any right angled triangle, write down what the equation is in terms of the sides of that triangle, and use the Pythagorean Identity to finish it off.

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Okay, Thanks a lot –  Sam Creamer Feb 8 '12 at 22:58

Just suppose that we have forgotten this very important property and try to evaluate the two integrals:

$$\int 1 dx = x+c$$

Is very trivial so we have now to proof that:

$$\int (\sin^2x+\cos^2x) dx=x+c$$

Now this kind of integral are usually explained in textbooks and you can find a few different ways to get that:

$$\int (\sin^2x)dx=\frac x2-\frac 14 \sin(2x)+c$$ $$\int (\cos^2x)dx=\frac x2+\frac 14 \sin(2x)+c$$

Now sum the two integral and you get the searched equalities.

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Any particular reason why the integration constant is the same in all three integrals? –  Calculon Jul 23 at 13:53
    
$c$ stands for "a real additive constant" so it doesn't matter to specify them with different letters. @Calculon –  Renato Faraone Jul 23 at 14:28

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