Substitution Rule for Definite Integrals — Question 9

Question 9

Evaluate the definite integral $$\int_{0}^{1} \frac{\sqrt{1-x}}{1+x}\,dx.$$

Question 9 - Solution

This integral becomes simpler if we eliminate the square root by substitution.

Let $$u=\sqrt{1-x}.$$ Then $$x=1-u^2, \qquad dx=-2u\,du.$$

Change the limits of integration. When $x=0$, $u=1$. When $x=1$, $u=0$.

Substitute into the integral: $$\int_{0}^{1} \frac{\sqrt{1-x}}{1+x}\,dx = \int_{1}^{0} \frac{u}{1+(1-u^2)}(-2u)\,du.$$

Simplify: $$\int_{1}^{0} \frac{-2u^2}{2-u^2}\,du = \int_{0}^{1} \frac{2u^2}{2-u^2}\,du.$$

Rewrite the integrand: $$\frac{2u^2}{2-u^2} = -2+\frac{4}{2-u^2}.$$

Thus, $$\int_{0}^{1} \left(-2+\frac{4}{2-u^2}\right)\,du.$$

Integrate term by term: $$\int -2\,du=-2u,$$ $$\int \frac{4}{2-u^2}\,du = 2\int \frac{1}{1-(u/\sqrt{2})^2}\,du.$$

Use the standard result $$\int \frac{1}{1-a^2}\,da=\operatorname{artanh}(a).$$

Thus, $$2\int \frac{1}{1-(u/\sqrt{2})^2}\,du = 2\sqrt{2}\,\operatorname{artanh}\!\left(\frac{u}{\sqrt{2}}\right).$$

Evaluate from $0$ to $1$: $$-2u+2\sqrt{2}\,\operatorname{artanh}\!\left(\frac{u}{\sqrt{2}}\right) \Big|_{0}^{1}.$$

Final answer: $$\boxed{ -2+2\sqrt{2}\,\operatorname{artanh}\!\left(\frac{1}{\sqrt{2}}\right) }$$