# Integration by Substitution

## Solved Problems

### Example 9.

Compute the integral $\int {\frac{{xdx}}{{1 + {x^4}}}}.$

Solution.

We can try the substitution $$u = {x^2}.$$ Then

$du = 2xdx,\;\; \Rightarrow xdx = \frac{{du}}{2}.$

Hence, the integral is equal to

$\int {\frac{{xdx}}{{1 + {x^4}}}} = \int {\frac{{\frac{{du}}{2}}}{{1 + {u^2}}}} = \frac{1}{2}\int {\frac{{du}}{{1 + {u^2}}}} = \frac{1}{2}\arctan u + C = \frac{1}{2}\arctan {x^2} + C.$

### Example 10.

Evaluate the integral $\int {\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}}.$

Solution.

If we complete the square in the denominator:

${x^4} + 2{x^2} + 1 = {\left( {{x^2} + 1} \right)^2} = {\left( {1 + {x^2}} \right)^2},$

we can use the substitution $$u = 1 + {x^2}.$$ Write the differential $$du:$$

$du = d\left( {1 + {x^2}} \right) = 2xdx.$

We see that the numerator can be expressed in terms of $$u:$$

$xdx = \frac{{du}}{2}.$

Hence, the integral is given by

$\int {\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}} = \int {\frac{{xdx}}{{{{\left( {1 + {x^2}} \right)}^2}}}} = \int {\frac{{\frac{{du}}{2}}}{{{u^2}}}} = \frac{1}{2}\int {\frac{{du}}{{{u^2}}}} = \frac{1}{2}\int {{u^{ - 2}}du} = \frac{1}{2} \cdot \frac{{{u^{ - 1}}}}{{\left( { - 1} \right)}} + C = - \frac{1}{{2u}} + C = - \frac{1}{{2\left( {1 + {x^2}} \right)}} + C.$

### Example 11.

Calculate the integral $\int {{2^x}{e^x}dx}.$

Solution.

We 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 12.

Find the integral $\int {x{e^{ - {x^2}}}dx}.$

Solution.

Using the substitution $$u = - {x^2},$$ we have

$du = d\left( { - {x^2}} \right) = - 2xdx.$

Note that

$xdx = - \frac{{du}}{2},$

so we can rewrite the integral in terms of the variable $$u$$ and solve it:

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

### Example 13.

Evaluate the integral $\int {\frac{{\sin x}}{{1 - \cos x}} dx}.$

Solution.

We make the substitution $$u = 1 - \cos x.$$ Hence

$du = - \left( { - \sin x} \right)dx = \sin xdx.$

This gives

$\int {\frac{{\sin x}}{{1 - \cos x}}dx} = \int {\frac{{du}}{u}} = \ln \left| u \right| + C = \ln \left| {1 - \cos x} \right| + C.$

### Example 14.

Evaluate the integral $\int {x\sqrt {x + 1} dx}.$

Solution.

To get rid of the square root, we make the substitution $$u = \sqrt {x + 1} .$$ Then

${u^2} = x + 1,\;\; \Rightarrow x = {u^2} - 1,\;\; \Rightarrow du = 2udu.$

The integral becomes

$\int {x\sqrt {x + 1} dx} = \int {\left( {{u^2} - 1} \right)u \cdot 2udu} = 2\int {\left( {{u^2} - 1} \right){u^2}du} = 2\int {\left( {{u^4} - {u^2}} \right)du} = 2\int {{u^4}du} - 2\int {{u^2}du} = 2 \cdot \frac{{{u^5}}}{5} - 2 \cdot \frac{{{u^3}}}{3} + C = \frac{2}{5}{\left( {x + 1} \right)^{\frac{5}{2}}} - \frac{2}{3}{\left( {x + 1} \right)^{\frac{3}{2}}} + C.$

### Example 15.

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

$u = \sin\left( {3x + 5} \right),\;\;\;du = 3\cos \left( {3x + 5} \right)dx,\;\; \Rightarrow \cos \left( {3x + 5} \right)dx = \frac{{du}}{3},$

$\int {\cot \left( {3x + 5} \right)dx} = \int {\frac{{\cos \left( {3x + 5} \right)}}{{\sin\left( {3x + 5} \right)}}dx} = \int {\frac{{\frac{{du}}{3}}}{u}} = \frac{1}{3}\int {\frac{{du}}{u}} = \frac{1}{3}\ln \left| u \right| + C = \frac{1}{3}\ln \left| {\sin\left( {3x + 5} \right)} \right| + C.$

### Example 16.

Find the integral $\int {{\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}} dx}.$

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,

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