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   Change of Variable
Let \(F\left( x \right)\) be an indefinite integral or antiderivative of \(f\left( x \right).\) Then \[ {\int {f\left( x \right)dx} } = {\int {f\left( {g\left( u \right)} \right)g'\left( u \right)du} } = {F\left( u \right) } = {F\left( {{g^{ - 1}}\left( x \right)} \right),} \] where \(x = g\left( u \right)\) is a substitution. Accordingly, the inverse function \(u = {g^{ - 1}}\left( x \right)\) describes the dependence of the new variable on the old variable.

It's important to remember that the differential \(dx\) also needs to be substituted. It must be replaced with the differential of the new variable \(du.\) For definite integrals, it is also necessary to change the limits of integration. See about this on the page "The Definite Integral and Fundamental Theorem of Calculus".

   Example 1
Calculate the integral \(\int {\large\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}\normalsize} .\)

Solution.
Let \(u = \large\frac{x}{a}\normalsize.\) Then \(x = au,\) \(dx = adu.\) Hence, the integral is \[\require{cancel} {\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} } = {\int {\frac{{adu}}{{\sqrt {{a^2} - {{\left( {au} \right)}^2}} }}} } = {\int {\frac{{adu}}{{\sqrt {{a^2}\left( {1 - {u^2}} \right)} }}} } = {\int {\frac{{\cancel{a}du}}{{\cancel{a}\sqrt {1 - {u^2}} }}} } = {\int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} } = {\arcsin u + C } = {\arcsin \frac{x}{a} + C.} \]
   Example 2
Find the integral \(\int {{\large\frac{{x + 1}}{{{x^2} + 2x - 5}}\normalsize} dx}. \)

Solution.
We make the substitution \(u = {x^2} + 2x - 5.\) Then \(du = 2xdx + 2 = 2\left( {x + 1} \right)dx\) or \(\left( {x + 1} \right)dx = {\large\frac{{du}}{2}\normalsize}.\)
The integral is easy to calculate with the new variable: \[ {\int {\frac{{x + 1}}{{{x^2} + 2x - 5}}dx} } = {\int {\frac{{\frac{{du}}{2}}}{u}} } = {\frac{1}{2}\int {\frac{{du}}{u}} } = {\frac{1}{2}\ln \left| u \right| + C } = {\frac{1}{2}\ln \left| {{x^2} + 2x - 5} \right| + C.} \]
   Example 3
Calculate the integral \(\int {{2^x}{e^x}dx} .\)

Solution.
Rewrite the integral in the following way: \[\int {{2^x}{e^x}dx} = \int {{{\left( {2e} \right)}^x}dx} .\] Denoting \(2e = a\) (this is not a change of variable, since \(x\) still remains the independent variable), we get the table integral: \[ {\int {{{\left( {2e} \right)}^x}dx} } = {\int {{a^x}dx} } = {\frac{{{a^x}}}{{\ln a}} + C } = {\frac{{{{\left( {2e} \right)}^x}}}{{\ln \left( {2e} \right)}} + C } = {\frac{{{2^x}{e^x}}}{{\ln 2 + \ln e}} + C } = {\frac{{{2^x}{e^x}}}{{\ln 2 + 1}} + C.} \]
   Example 4
Calculate the integral \(\int {\cot \left( {3x + 5} \right)dx}.\)

Solution.
We can write the integral as \[ {\int {\cot \left( {3x + 5} \right)dx} } = {\int {\frac{{\cos \left( {3x + 5} \right)}}{{\sin\left( {3x + 5} \right)}}dx} .} \] Changing the variable \[ {u = \sin\left( {3x + 5} \right),}\;\; {du = 3\cos \left( {3x + 5} \right)dx,}\;\; {\Rightarrow \cos \left( {3x + 5} \right)dx = \frac{{du}}{3},} \] we obtain the answer \[ {\int {\cot \left( {3x + 5} \right)dx} } = {\int {\frac{{\cos \left( {3x + 5} \right)}}{{\sin\left( {3x + 5} \right)}}dx} } = {\int {\frac{{\frac{{du}}{3}}}{u}} } = {\frac{1}{3}\int {\frac{{du}}{u}} } = {\frac{1}{3}\ln \left| u \right| + C } = {\frac{1}{3}\ln \left| {\sin\left( {3x + 5} \right)} \right| + C.} \]
   Example 5
Find the integral \(\int {{\large\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}\normalsize} dx}.\)

Solution.
We make the following substitution: \[ {u = 1 + {\cos ^2}x,}\;\; {\Rightarrow du = {\left( {1 + {{\cos }^2}x} \right)^\prime }dx } = {2\cos x \cdot \left( { - \sin x} \right)dx } = { - \sin 2xdx.} \] Hence, \[ {\int {\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}dx} } = {\int {\frac{{\left( { - du} \right)}}{{\sqrt u }}} } = { - 2\int {\frac{{du}}{{2\sqrt u }}} } = { - 2\sqrt u + C } = { - 2\sqrt {1 + {{\cos }^2}x} + C.} \]

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