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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, ${\int {\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}dx} } = {\int {\frac{{\left( { - du} \right)}}{{\sqrt u }}} } = { - 2\int {\frac{{du}}{{2\sqrt u }}} } = { - 2\sqrt u + C } = { - 2\sqrt {1 + {{\cos }^2}x} + C.}$