Definite Integrals Using Substitution (Class XII, Ex 7.10)

This lesson works through four definite integrals from Class XII Exercise 7.10, each solved by a suitable substitution. For each one we choose a substitution, transform the integral, and evaluate carefully, paying attention to how the limits are handled.

Example 1: a fifth-power expression under a root

Evaluate

$\int_0^1 5x^4\sqrt{x^5+1}\,dx.$

Put $t = x^5 + 1$, so $dt = 5x^4\,dx$.

Method 1 (substitute back). The integral becomes $\int \sqrt{t}\,dt = \tfrac{2}{3}t^{3/2}$. Returning to $x$ gives

$\tfrac{2}{3}\left(x^5+1\right)^{3/2}\Big|_0^1 = \tfrac{2}{3}\left(2^{3/2} - 1^{3/2}\right) = \tfrac{2}{3}\left(2\sqrt{2} - 1\right).$

Method 2 (change the limits). When $x = 0$, $t = 1$; when $x = 1$, $t = 2$. So

$\int_1^2 \sqrt{t}\,dt = \tfrac{2}{3}t^{3/2}\Big|_1^2 = \tfrac{2}{3}\left(2\sqrt{2} - 1\right).$

Both methods agree.

Example 2: a square-root term set equal to $t^2$

Evaluate

$\int_0^2 x\sqrt{x+2}\,dx.$

The key step is to put $x + 2 = t^2$. Then $dx = 2t\,dt$ and $x = t^2 - 2$. Changing the limits: when $x = 0$, $t = \sqrt{2}$; when $x = 2$, $t = 2$. The integral becomes

$\int_{\sqrt{2}}^{2} \left(t^2 - 2\right)\cdot t \cdot 2t\,dt = 2\int_{\sqrt{2}}^{2}\left(t^4 - 2t^2\right)dt.$

Integrating,

$2\left[\frac{t^5}{5} - \frac{2t^3}{3}\right]_{\sqrt{2}}^{2}.$

The rational part gives $\tfrac{32}{5} - \tfrac{16}{3} = \tfrac{16}{15}$, and the $\sqrt{2}$ part gives $\tfrac{4\sqrt{2}}{5} - \tfrac{4\sqrt{2}}{3} = -\tfrac{8\sqrt{2}}{15}$. Combining,

$2\left(\frac{16}{15} + \frac{8\sqrt{2}}{15}\right) = \frac{32 + 16\sqrt{2}}{15} = \frac{16\sqrt{2}}{15}\left(\sqrt{2} + 1\right).$

Example 3: an exponential factor

Evaluate

$\int_1^2 \left(\frac{1}{x} - \frac{1}{2x^2}\right)e^{2x}\,dx.$

Put $2x = t$, so $2\,dx = dt$ and $x = \tfrac{t}{2}$. When $x = 1$, $t = 2$; when $x = 2$, $t = 4$. After substituting and simplifying, the integrand takes the form

$\int_2^4 \left(\frac{1}{t} - \frac{1}{t^2}\right)e^{t}\,dt.$

This fits the pattern $\int e^{t}\left(f(t) + f'(t)\right)dt = e^{t}f(t)$ with $f(t) = \tfrac{1}{t}$, since $f'(t) = -\tfrac{1}{t^2}$. Hence

$\frac{1}{t}e^{t}\Big|_2^4 = \frac{e^4}{4} - \frac{e^2}{2} = \frac{e^2\left(e^2 - 2\right)}{4}.$

Example 4: a reciprocal substitution (objective type)

Evaluate

$\int_{1/3}^{1} \frac{\left(x - x^3\right)^{1/3}}{x^4}\,dx.$

Multiply and divide so the cube of $x$ is pulled inside the cube root: $\left(x - x^3\right)^{1/3} = x\left(\tfrac{1}{x^2} - 1\right)^{1/3}$. The integral becomes

$\int_{1/3}^{1} \frac{\left(\tfrac{1}{x^2} - 1\right)^{1/3}}{x^3}\,dx.$

Put $t = \dfrac{1}{x^2}$. Then $\dfrac{1}{x^3}\,dx = -\tfrac{1}{2}\,dt$. Changing the limits: when $x = \tfrac{1}{3}$, $t = 9$; when $x = 1$, $t = 1$. So

$-\tfrac{1}{2}\int_{9}^{1}\left(t - 1\right)^{1/3}\,dt = -\tfrac{1}{2}\cdot \frac{3}{4}\left(t - 1\right)^{4/3}\Big|_{9}^{1}.$

Evaluating,

$-\tfrac{3}{8}\left(0 - 8^{4/3}\right) = -\tfrac{3}{8}\left(-16\right) = 6.$

The answer is option (a), $6$.

Key takeaways

• Choose a substitution that makes the awkward part of the integrand simple, often the term inside a root or power.
• You can either substitute back to the original variable before applying the given limits, or change the limits to the new variable and finish there.
• Setting a square-root term equal to $t^2$ clears the radical and turns the integral into a polynomial.
• Recognising the form $\int e^{t}\left(f + f'\right)dt = e^{t}f$ shortens exponential integrals dramatically.