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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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    But you'll probably make your teacher happier if you put in some parentheses: $\int(\sin^2x+\cos^2x)\,dx=\int1\,dx$.2012-02-08

2 Answers 2

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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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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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    $c$ stands for "a real additive constant" so it doesn't matter to specify them with different letters. @Calculon2015-07-23