U-substitution to solve definite integrals
U-substitution in definite integrals is a little different than substitution in indefinite integrals
U-substitution in definite integrals is just like substitution in indefinite integrals except that, since the variable is changed, the limits of integration must be changed as well.
When using u-substitution in definite integrals, you either have to change the limits of integration, or back-substitute
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Evaluating a definite integral using u-substitution
Example
Use u-substitution to evaluate the integral.
???\int_0^{\frac{\pi}{2}}\frac{\cos{x}}{1+\sin^2{x}}\ dx???
Let
???u=\sin{x}???
???du=\cos{x}\ dx???
???dx=\frac{du}{\cos{x}}???
Since we’re dealing with a definite integral, we need to use the equation ???u=\sin{x}??? to find limits of integration in terms of ???u???, instead of ???x???.
Substituting back into the integral (including for our limits of integration), we get
???\int_0^1\frac{\cos{x}}{1+u^2}\cdot\frac{du}{\cos{x}}???
???\int_0^1\frac{1}{1+u^2}\ du???
Using this very common formula,
???\int\frac{1}{1+x^2}\ dx=\tan^{-1}{x}+C???
take the integral.
???\int_0^1\frac{1}{1+u^2}\ du=\tan^{-1}{u}\big|_0^1???
???\int_0^1\frac{1}{1+u^2}\ du=\tan^{-1}{1}-\tan^{-1}{0}???
???\int_0^1\frac{1}{1+u^2}\ du=\frac{\pi}{4}???