**Example 2:**

Integrate the following functions using the substitution method:

**a)** $\frac{\int\ tan\ x}{sec^4\ x\ dx}$

**b)** $\frac{\int\ dx}{1 + e^x\ dx}$

**c)** $\int\ 2x\ (x^2 + 1)^3\ dx$

**Solution: **

**a)** $\frac{\int\ tan\ x}{sec^4\ x\ dx}$

It is possible to apply integration by substitution when the integral contains both the function and its derivative. So, to get that form we multiply both the numerator and denominator by sec $x$.

$\frac{\int\ sec\ x\ tan\ x}{sec^5\ x\ dx}$

Let us substitute sec $x$ = $u$, then on differentiating we get sec $x$ tan $x\ dx$ = $du$. The integral now looks like,

= $\frac{\int\ du}{u^5}$

= $\int\ u^{-5}\ du$

= $u^{\frac{(-5 + 1)}{(-5 + 1)}}$ + $c$

= $u^{\frac{-4}{-4}}$ + $c$

= $\frac{- 1}{4u^4}$ + $c$

Plugging in back the value of $u$ in the solution we get,

= $\frac{- 1}{sec^4}$ $x + c$

**b)** $\frac{\int\ dx}{1}$ + $e^x\ dx$

It is possible to apply integration by substitution when the integral contains both the function and its derivative. So, to get that form we multiply both the numerator and denominator by $e^x$

$\frac{\int\ e^x\ dx}{e^x(1 + e^x)\ dx}$

Let us substitute $e^x$ = $u$, then on differentiating we get $e^x\ dx$ = $du$. The integral now looks like,

= $\frac{\int\ du}{u(1 + u)}$

= $\frac{\int\ (1 + u)\ –\ u\ du}{u(1 + u)}$

= $\frac{\int\ du}{u}$ - $\frac{\int\ du}{(1 + u)}$

= $ln |u| - ln |1 + u| + c$

= $ln$ $|\frac{u}{1}$ + $u|\ +\ c$

Plugging in back the value of $u$ in the solution we got,

= $ln$ $|\frac{e^x}{1}$ + $e^x|\ +\ c$

**c)** $\int\ 2x\ (x^2 + 1)^3\ dx$

Let us substitute $x^2 + 1$ = $u$, then on differentiating we get $2x\ dx$ = $du$

= $\int\ u^3\ du$

= $u^{\frac{(3 + 1)}{(3 + 1)}}$ + $c$

= $\frac{u^4}{4}$ + $c$

= $\frac{(x^2 + 1)}{4}$ + $c$