Integration by substitution

The substitution method, in general, works whenever we have an integral that we can write in the form ∫ f(g(x)) g'(x) dx. Observe that if F' = f, then

because, by the Chain Rule

If we make the "change of variable" or substition u = g(x), then we have

since F' = f

The Substitution Rule for definite integrals states:

$${\int }_a^{b}f(g(x))g^{\prime }(x)\text{d}x={\int }_{g(a)}^{g(b)}f(u)\text{d}u$$

If g'(x) is a continuous function on [a,b] and if f is continuous on the range of g, then

Example 2. Caluclate the following integral by substitution:

Solution. The substitution t = six, yields:

since cos x dx = dt. Substituting back t, we have:

Example 1 Caluclate the following definite integral by substitution:

$${\int }_0^1\sqrt{1-x^2}\text{d}x$$

the substitution x = cos t, yields

$$\begin{array}{rl}{\int }_0^1\sqrt{1-x^2}dx& ={\int }_0^{\pi /2}\sqrt{1-{\sin }^{2}t}\cos (t)dt\\ & ={\int }_0^{\pi /2}{\cos }^{2}tdt\\ & =[\frac{t}{2}+\frac{\sin (2t)}{4}]{|}_0^{\pi /2}\\ & =\frac{\pi }{4}+0=\frac{\pi }{4}.\end{array}$$

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