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Explore Stack InternalAnother neat trick is to add different forms of integrals to obtain a much simpler one.
For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=\int_{-1}^1\frac1{1+f(x)}\,dx$$ then we could replace $x$ by $-x$ giving $$I=-\int_1^{-1}\frac1{1+f(-x)}\,dx=\int_{-1}^1\frac{f(x)}{1+f(x)}\,dx$$ and adding gives $$2I=\int_{-1}^1\,dx.$$$$2I=\int_{-1}^1\,dx=2\implies I=1.$$
Another neat trick is to add different forms of integrals to obtain a much simpler one.
For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=\int_{-1}^1\frac1{1+f(x)}\,dx$$ then we could replace $x$ by $-x$ giving $$I=-\int_1^{-1}\frac1{1+f(-x)}\,dx=\int_{-1}^1\frac{f(x)}{1+f(x)}\,dx$$ and adding gives $$2I=\int_{-1}^1\,dx.$$
Another neat trick is to add different forms of integrals to obtain a much simpler one.
For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=\int_{-1}^1\frac1{1+f(x)}\,dx$$ then we could replace $x$ by $-x$ giving $$I=-\int_1^{-1}\frac1{1+f(-x)}\,dx=\int_{-1}^1\frac{f(x)}{1+f(x)}\,dx$$ and adding gives $$2I=\int_{-1}^1\,dx=2\implies I=1.$$
Another neat trick is to add different forms of integrals to obtain a much simpler one.
For example, if we let a function $f$ be such that $f(x)f(-x)=1$ and we want to evaluate $$I=\int_{-1}^1\frac1{1+f(x)}\,dx$$ then we could replace $x$ by $-x$ giving $$I=-\int_1^{-1}\frac1{1+f(-x)}\,dx=\int_{-1}^1\frac{f(x)}{1+f(x)}\,dx$$ and adding gives $$2I=\int_{-1}^1\,dx.$$
