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$$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin^2 x dx$$

My way of solving -

$$2 \int_0^{\frac{\pi}{4}} \frac{1- \cos2x}{2} dx$$

On solving i get $\dfrac{\pi}4 - \dfrac 12$ as an answer.

My main question is can we us any other property here also?

Something like -

$$2 \int_0^{\frac{\pi}{4}} \sin^2x dx$$

$$2 \int_0^{\frac{\pi}{4}} \sin^2(\frac{\pi}{4}-x)dx$$

Then how to proceed?

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    The first approach is clearly much quicker as in the second one you might have to get in a lot of trouble with trigonometric identities to get to the same result, unless you approach it graphically. – 2017-02-11

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The method you suggest is the best, but if you really want a different method, you could try integration by parts to get $$I=-\sin x\cos x+\int \cos^2 x dx$$ and then use $\cos^2 x =1-\sin^2 x$ to get an expression for $2I$ And so on...