Etydhv

Let $$I=int fracsin xcos x + sin x dx$$

Now let $$u=fracpi2 - x$$ so $$I=int frac-sin (fracpi2 - u)cos (fracpi2 - u)+sin (fracpi2 - u) du$$

$$=intfrac-cos usin u + cos u du$$

$$= intfrac-cos xsin x + cos x dx$$

and hence $$2I=intfracsin x - cos xsin x + cos x dx$$

$$=-ln |sin x + cos x| + c$$

$implies I=-frac12ln |sin x + cos x| + c$

But the actual answer is $$I= frac12x -frac12ln |sin x + cos x| + c$$

according to Wolfram Alpha and verified by a different method.

Why does my method not yield the correct result?

You have not paid enough attention to the limits of your integration - the two integrals you are adding are not over the same interval.

The method to use here is quite similar to yours in using the same trick correctly, without disturbing the interval of integration:

$$frac2sin xsin x + cos x=frac sin x +cos xsin x +cos x+frac sin x - cos xsin x + cos x$$

You might need a little more explanation of the interval issue.

Integrals are additive in two ways.

First, if the same function is integrated over disjoint intervals (later measurable sets) then we can integrate over the union of the intervals and the integral over the whole is the sum of the integrals over the parts.

Second, if we integrate different functions over the same interval (measurable set), the sum of the integrals is equal to the integral of the sum of the functions.

Apply your method to the integral of $x^2$ and use the substitution $y=-x$. The integral you get is the integral of $-x^2$. Adding the two you get twice the integral is zero, which is nonsense, because the function you are integrating is positive except at $x=0$. What has happened here is that you have swapped the limits of integration, and you need to reverse the sign to straighten them out.

Your method has both reversed the limits and translated them by $frac pi 2$. You can't add the integrals in this case and expected to get the right answer.

Going by the logic you have used, if we subtract the two equations, we'll get:

$$0 = int 1 dx $$

Which is certainly wrong.

$$int_a^bfracsin xcos x + sin xdxneq int_a^bfrac-cos xcos x + sin xdx$$
but
$$int_a^bfracsin xcos x + sin xdx= int_pi/2-a^pi/2-bfrac-cos xcos x + sin xdx$$

The incorrect part is assuming: $$intfrac-cos usin u + cos u du=intfrac-cos xsin x + cos x dx$$
One is depending on $x$ while the other one depends on $u$.

Integrating an indefinite integral it's like having a function that depends on a parameter, and we can look at it as the opposite of the derivative (not as computing the surface under some function).

Yes, for a definite integral we have:$$int_a^b frac-cos usin u + cos u du=int_a^bfrac-cos xsin x + cos x dx$$
But that is because the definite integral computes the area and the result will always be a constant ($x$ and $u$ are dummy variables in this case).