The method of substitution is one of the basic methods for calculating indefinite integrals. Even when we integrate by some other technique, we often resort to substitution in the intermediate stages. The success of integration largely depends on how appropriate the substitution is, in simplifying the integrand. Essentially, the study of the methods of integration, reduces to finding out what kind of substitution should be done. I solve a few interesting problems here.

##### Change of variable

Theorem. The form of the integration formula is independent of the nature of the variable of integration.

Proof. Let it be required to find the integral

$$\int{f(x)dx}$$

Let us change the variable in the expression under the integral sign, putting

$$x=\phi(t)$$

$$\int{f(x)dx}=\int{f[\phi(t)]d(\phi(t))}=\int{f[\phi(t)]\phi'(t)dt}$$

In many problems, a substitution such as the one above, leads to a simpler integral. Let us establish, that the indefinite integral is independent of the nature of the variable of integration – whether $$x$$ or $$\phi(t)$$. It is necessary to prove that their derivatives with respect to $$x$$ are equal.

Differentiating the left side with respect to $$x$$:

$$\left(\int{f(x)dx}\right)_{x}=f(x)$$

Differentiating the right side with respect to $$x$$:

$$\left(\int{f[\phi(t)]\phi'(t)dt}\right)_{x}=\left(\int{f[\phi(t)]\phi'(t)dt}\right)_{t}\cdot\frac{dt}{dx}=f[\phi(t)]\phi'(t)\frac{dt}{dx}$$

But,

$$\frac{dt}{dx}=\frac{1}{dx/dt}=\frac{1}{\phi'(t)}$$

Therefore,

$$\left(\int{f[\phi(t)]\phi'(t)dt}\right)_{x}=f[\phi(t)]\phi'(t)\frac{1}{\phi'(t)}=f[\phi(t)]=f(x)$$

Therefore, the derivatives, with repsect to $$x$$ of the right and left sides are equal as required. Hence, the expressions to the right and left side are the same.

##### Examples

Example. \begin{aligned} \int{\frac{dx}{\cos^{2}x(3\tan{x}+1)}}\\ \end{aligned}

Solution. We make the substitution:

$$3\tan{x}+1=t$$.

$$\displaystyle{\frac{3dx}{\cos^{2}{x}}=dt}$$.

\begin{aligned} I&=\frac{1}{3}\int{\frac{dt}{t}}\\ &=\frac{1}{3}\ln{(t)}+C\\ &=\frac{1}{3}\ln{(3\tan{x}+1})+C \end{aligned}

Example. \begin{aligned} \int{\frac{dx}{9x^{2}+4}}\\ \end{aligned}

Solution. We make the substitution:

$$3x=t$$.

$$3dx=dt$$.

\begin{aligned} I&=\frac{1}{3}\int{\frac{dt}{2^{2}+t^{2}}}\\ &=\frac{1}{6}\arctan{\left(\frac{t}{2}\right)}+C\\ &=\frac{1}{6}\arctan{\left(\frac{3x}{2}\right)}+C \end{aligned}

Example. \begin{aligned} \int{\frac{dx}{4-9x^{2}}}\\ \end{aligned}

Solution. We make the substitution $$3x=t$$ as before.

\begin{aligned} I&=\frac{1}{3}\int{\frac{dt}{2^{2}-t^{2}}}\\ &=\frac{1}{3}\cdot\frac{1}{2}\ln{\left(\frac{2+t}{2-t}\right)}+C\\ &=\frac{1}{6}\ln{\left(\frac{2+3x}{2-3x}\right)}+C\\ \end{aligned}

Example. \begin{aligned} \int{\frac{dx}{\sqrt{b^{2}x^{2}-a^{2}}}}\\ \end{aligned}

Solution. We make the substitution $$bx=t$$.

\begin{aligned} I&=\frac{1}{b}\int{\frac{dt}{\sqrt{t^{2}-a^{2}}}}\\ &=\frac{1}{b}\ln{(t+\sqrt{t^{2}-a^{2}})}+C\\ &=\frac{1}{b}\ln{(bx+\sqrt{b^{2}x^{2}-a^{2}})}+C\\ \end{aligned}

Example. \begin{aligned} \int{\frac{x^{2}dx}{5-x^{6}}}\\ \end{aligned}

Solution. We make the substitution $$x^{3}=t$$.

\begin{aligned} I&=\frac{1}{3}\int{\frac{dt}{(\sqrt{5})^2-t^{2}}}\\ &=\frac{1}{6\sqrt{5}}\ln\left(\frac{\sqrt{5}+t}{\sqrt{5}-t}\right)+C\\ &=\frac{1}{6\sqrt{5}}\ln\left(\frac{\sqrt{5}+x^{3}}{\sqrt{5}-x^{3}}\right)+C\\ \end{aligned}

Example. \begin{aligned} \int{\frac{xdx}{\sqrt{1-x^{4}}}}\\ \end{aligned}

Solution. We make the substitution $$x^{2}=t$$.

\begin{aligned} I&=\frac{1}{2}\int{\frac{dt}{\sqrt{1-t^{2}}}}\\ &=\frac{1}{2}\arcsin{(t)}+C\\ &=\frac{1}{2}\arcsin(x^{2})+C\\ \end{aligned}

Example. \begin{aligned} \int{\frac{\arccos{x}-x}{\sqrt{1-x^{2}}}dx}\\ \end{aligned}

Solution. The indefinite integral can be written as:

\begin{aligned} I&=\int{\frac{\arccos(x)}{\sqrt{1-x^{2}}}dx}-\int{\frac{x{dx}}{\sqrt{1-x^{2}}}}\\ &=I_{1}-I_{2}\\ \end{aligned}

In the first integral, we make the substitution $$\arccos(x)=t$$.

\begin{aligned} I_{1}&=-\int{tdt}\\ &=-\frac{t^{2}}{2}+C_{1}\\ &=-\frac{\arccos^{2}x}{2}+C_{1} \end{aligned}

In the second integral, we make the substitution $$1-x^{2}=t$$.

\begin{aligned} I_{2}&=-\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}\\ &=-\frac{1}{2}2\sqrt{t}+C_{2}\\ &=-\sqrt{1-x^{2}}+C_{2} \end{aligned}

The desired indefinite integral is

\begin{aligned} I&=-\frac{\arccos^{2}x}{2}+\sqrt{1-x^{2}}+C \end{aligned}

Example. \begin{aligned} \int{\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x}}dx} \end{aligned}

Solution. We make the substitution $$1+\sqrt{x}=t$$.

\begin{aligned} I&=2\int{\sqrt{t}dt}\\ &=2\frac{t^{3/2}}{3/2}+C\\ &=\frac{4}{3}t^{3/2}+C\\ &=\frac{4}{3}(1+\sqrt{x})^{3/2}+C \end{aligned}

Example. \begin{aligned} \int{\frac{dx}{2\sin^{2}{x}+3\cos^{2}{x}}} \end{aligned}

Solution. We divide throughout by $$\cos^{2}{x}$$. Consequently, we have :

\begin{aligned} I&=\int{\frac{\sec^{2}{x}dx}{2\tan^{2}x+3}} \end{aligned}

We make the substitution $$\sqrt{2}\tan{x}=t$$.

\begin{aligned} I&=\frac{1}{\sqrt{2}}\int{\frac{dt}{t^{2}x+(\sqrt{3})^{2}}}\\ &=\frac{1}{\sqrt{6}}\arctan\left(\frac{t}{\sqrt{3}}\right)+C\\ &=\frac{1}{\sqrt{6}}\arctan\left(\sqrt{\frac{2}{3}}x\right)+C\\ \end{aligned}

Example. \begin{aligned} \int{\sqrt{\frac{1-x}{1+x}}dx} \end{aligned}

Solution. We multiply the numerator and denominator by $$\sqrt{1-x}$$. Consequently, we have :

\begin{aligned} I&=\int{\frac{1-x}{\sqrt{1-x^{2}}}dx}\\ &=\int{\frac{dx}{\sqrt{1-x^{2}}}}-\int{\frac{xdx}{\sqrt{1-x^{2}}}}\\ &=\arcsin{x}+\frac{1}{2}\int{\frac{d(-x^{2})}{\sqrt{1-x^{2}}}}\\ &=\arcsin{x}+\frac{1}{2}\cdot{2}\sqrt{1-x^{2}}+C\\ &=\arcsin{x}+\sqrt{1-x^{2}}+C\\ \end{aligned}