Integration by substitution is a rule of integration which is also known as \(u\)-substitution or change of variables.

The substitution rule/ \(u\)-substitution: If \(f\) is a continuous function on an interval \(I\) and \(u=g(x)\) is a differentiable function whose range is inside \(I\) where \(g'\) is continuous, then \[\displaystyle\int f(g(x)) g'(x) \,dx= \int f(u) \,du. \]

Example. Evaluate \(\displaystyle\int xe^{x^2} \,dx\).
Solution. Let \(u=x^2\). Then \(du=2x dx\) and \(x dx=\frac{1}{2}du\). \[\begin{align*} \int xe^{x^2} \,dx &= \int e^{x^2} x\,dx &\\ &= \int e^u \frac{1}{2}du & (\text{By the $u$-substitution) }\\ &= \frac{1}{2}\int e^u \,du & \\ &= \frac{e^u}{2}+C &\\ &= \frac{e^{x^2}}{2}+C & \end{align*}\]

Example. Evaluate \(\displaystyle\int \tan x \,dx\).
Solution. \[\begin{align*} \int \tan x \,dx &= \int \frac{\sin x}{\cos x}\,dx &\\ &= \int \frac{1}{u} (-du) &(\text{Let } u=\cos x. \text{ Then } du=-\sin x \,dx\implies \sin x \,dx=-du )\\ &= -\int \frac{1}{u} \,du & \\ &= -\ln|u|+C & \\ &= \ln \left(\frac{1}{|u|} \right) +C & \\ &= \ln \left(\frac{1}{|\cos x|} \right) +C & \\ &= \ln (|\sec x|) +C & \end{align*}\]

Example. Evaluate \(\displaystyle\int_0^{\sqrt{5}} \frac{2x}{\sqrt{x^2+4}} \,dx\).
Solution. First we do the corresponding indefinite integral: \[\begin{align*} \int \frac{2x}{\sqrt{x^2+4}} \,dx &= \int \frac{du}{\sqrt{u}} &(\text{Let } u=x^2+4. \text{ Then } du=2x \,dx )\\ &= 2\sqrt{u} +C & \\ &= 2\sqrt{x^2+4} +C & \end{align*}\] Then \[ \displaystyle\int_0^{\sqrt{5}} \frac{2x}{\sqrt{x^2+4}} \,dx = \left. 2\sqrt{x^2+4} \right\vert_0^{\sqrt{5}} =2\sqrt{(\sqrt{5})^2+4}-2\sqrt{4}=6-4=2.\]

The preceding definite integral can also be evaluated by the following substitution rule/ \(u\)-substitution for definite integrals:
\[\displaystyle\int_a^b f(g(x)) g'(x) \,dx= \int_{g(a)}^{g(b)} f(u) \,du. \] Note that when \(x=a\), \(u=g(a)\) and when \(x=b\), \(u=g(b)\).

Example. Evaluate \(\displaystyle\int_0^{\sqrt{5}} \frac{2x}{\sqrt{x^2+4}} \,dx\).
Solution. We substitute \(u=x^2+4\) as before. Then \(du=2x \,dx\). When \(x=0\), \(u=4\) and when \(x=\sqrt{5}\), \(u=9\). \[\begin{align*} \int_0^{\sqrt{5}} \frac{2x}{\sqrt{x^2+4}} \,dx &= \int_4^9 \frac{du}{\sqrt{u}} \\ &= \left. 2\sqrt{u} \right\vert_4^9 \\ &= 2\sqrt{9}-2\sqrt{4} \\ &= 2 \end{align*}\]

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