Evaluation of Definite Integrals by Substitution

Substitution Meets Limits

Substitution was our workhorse for indefinite integrals. For a definite integral $\int_a^b f(x)\,dx$, NCERT first lists the careful four-step route:

1. Consider the integral without limits and substitute ($t = g(x)$ or $x = g(t)$) to reduce it to a known form.
2. Integrate in the new variable, without the constant of integration.
3. Resubstitute to return the answer to the original variable $x$.
4. Evaluate that answer at the given limits and take (value at upper limit) minus (value at lower limit).

The quicker route — move the limits with the variable

NCERT's Note: steps 1 and 2 stay, but step 3 disappears. Keep the integral in the new variable, and change the limits to match: if $t = g(x)$, then the $x$-limits $a, b$ become the $t$-limits $g(a), g(b)$.

$$\int_a^b f\big(g(x)\big)g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt$$

Worked contrast (NCERT Example 26): for $\int_{-1}^{1} 5x^4\sqrt{x^5 + 1}\,dx$ with $t = x^5 + 1$, $dt = 5x^4 dx$:

1. Long way: $\int\sqrt t\,dt = \frac23 t^{3/2}$, resubstitute to $\frac23(x^5+1)^{3/2}$, then evaluate at $\pm1$: $\frac23\left[2^{3/2} - 0\right] = \frac{4\sqrt2}{3}$.
2. Quick way: when $x = -1$, $t = 0$; when $x = 1$, $t = 2$. So the integral is $\int_0^2\sqrt t\,dt = \left[\frac23 t^{3/2}\right]_0^2 = \frac{4\sqrt2}{3}$ — same answer, no return trip.

Key Point: In a definite integral, the substitution changes three things at once — the integrand, the differential, and the limits. Once all three are moved to the $t$-world, finish there. Never mix: $t$-antiderivative with $x$-limits is the classic wrong answer.

Checklist before you finish

1. Did every $x$ leave the integral? After substituting, no stray $x$ may remain — if it does, the substitution isn't complete.
2. Did the limits move? Compute $t$ at $x = a$ and at $x = b$ and write them on the new integral sign immediately.
3. Lower limit can exceed upper. If $g(a) > g(b)$, leave them in that order — the sign sorts itself out. Do not "fix" the order.

[JEE Tip] The change-of-limits form is also how JEE expects you to read definite integrals backwards: recognising $\int_0^{\pi/4}\tan^3 t\,\sec^2 t\,dt$ as $\int_0^1 u^3\,du = \frac14$ in one line (with $u = \tan t$) is a routine time-saver. Also remember $\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x$ for $|x| \leq 1$ — substitution-friendly identities convert scary integrands into polynomial ones.

Solved Examples

Example 1: The two routes side by side

Evaluate $\displaystyle\int_{-1}^{1} 5x^4\sqrt{x^5 + 1}\,dx$.

Solution:

1. Substitute: $t = x^5 + 1$, $dt = 5x^4\,dx$.
2. Move the limits: $x = -1 \Rightarrow t = (-1)^5 + 1 = 0$; $x = 1 \Rightarrow t = 2$.
3. Integrate in $t$: $\int_0^2 \sqrt t\,dt = \left[\frac{2}{3}t^{3/2}\right]_0^2 = \frac{2}{3}\cdot 2\sqrt 2$.

Final Answer: $\dfrac{4\sqrt 2}{3}$.

Example 2: An inverse-trig pair

Evaluate $\displaystyle\int_0^1 \frac{\tan^{-1}x}{1 + x^2}\,dx$.

Solution:

1. Substitute: $t = \tan^{-1}x$, $dt = \frac{dx}{1 + x^2}$.
2. Move the limits: $x = 0 \Rightarrow t = 0$; $x = 1 \Rightarrow t = \frac{\pi}{4}$.
3. Integrate: $\int_0^{\pi/4} t\,dt = \left[\frac{t^2}{2}\right]_0^{\pi/4} = \frac{1}{2}\cdot\frac{\pi^2}{16}$.

Final Answer: $\dfrac{\pi^2}{32}$.

Takeaway: A function next to its own derivative ($\tan^{-1}x$ beside $\frac{1}{1+x^2}$) is the signature of substitution — the integral collapses to $\int t\,dt$.

Example 3: Log limits

Evaluate $\displaystyle\int_0^1 \frac{x}{x^2 + 1}\,dx$.

Solution:

1. Substitute: $t = x^2 + 1$, $dt = 2x\,dx$, so $x\,dx = \frac{dt}{2}$.
2. Move the limits: $x = 0 \Rightarrow t = 1$; $x = 1 \Rightarrow t = 2$.
3. Integrate: $\frac{1}{2}\int_1^2\frac{dt}{t} = \frac{1}{2}\left[\log t\right]_1^2 = \frac{1}{2}(\log 2 - 0)$.

Final Answer: $\dfrac{1}{2}\log 2$.

Example 4: Trig with reversed limits

Evaluate $\displaystyle\int_0^{\pi/2} \frac{\sin x}{1 + \cos^2 x}\,dx$.

Solution:

1. Substitute: $t = \cos x$, $dt = -\sin x\,dx$.
2. Move the limits: $x = 0 \Rightarrow t = 1$; $x = \frac{\pi}{2} \Rightarrow t = 0$ — the limits come out reversed; keep them.
3. Integrate: $\int_1^0 \frac{-dt}{1 + t^2} = \int_0^1\frac{dt}{1 + t^2} = \left[\tan^{-1}t\right]_0^1$.

Final Answer: $\dfrac{\pi}{4}$.

Takeaway: The minus sign from $dt = -\sin x\,dx$ and the reversed limits cancel each other. Handle both mechanically — flip the limits and absorb the sign — rather than guessing.

Example 5: Completing the square first

Evaluate $\displaystyle\int_{-1}^{1} \frac{dx}{x^2 + 2x + 5}$.

Solution:

1. Complete the square: $x^2 + 2x + 5 = (x + 1)^2 + 4$.
2. Substitute: $t = x + 1$, $dt = dx$; limits: $x = -1 \Rightarrow t = 0$, $x = 1 \Rightarrow t = 2$.
3. Integrate: $\int_0^2\frac{dt}{t^2 + 2^2} = \frac{1}{2}\left[\tan^{-1}\frac{t}{2}\right]_0^2 = \frac{1}{2}\left[\tan^{-1}1 - 0\right]$.

Final Answer: $\dfrac{\pi}{8}$.

Example 6: Powers of sine and cosine

Evaluate $\displaystyle\int_0^{\pi/2} \sqrt{\sin\phi}\,\cos^5\phi\;d\phi$.

Solution:

1. Save one cosine, convert the rest: $\cos^5\phi = \left(1 - \sin^2\phi\right)^2\cos\phi$.
2. Substitute: $t = \sin\phi$, $dt = \cos\phi\,d\phi$; limits $0 \to 0$, $\frac{\pi}{2} \to 1$.
3. Integrate: $\int_0^1 \sqrt t\,(1 - t^2)^2\,dt = \int_0^1\left(t^{1/2} - 2t^{5/2} + t^{9/2}\right)dt = \frac{2}{3} - \frac{4}{7} + \frac{2}{11}$.

Final Answer: $\dfrac{64}{231}$.

Example 7: A root substitution

Evaluate $\displaystyle\int_0^2 x\sqrt{x + 2}\,dx$ (put $x + 2 = t^2$).

Solution:

1. Substitute as instructed: $x = t^2 - 2$, $dx = 2t\,dt$; limits: $x = 0 \Rightarrow t = \sqrt2$, $x = 2 \Rightarrow t = 2$.
2. Rewrite: $\int_{\sqrt2}^{2}(t^2 - 2)\,t\cdot 2t\,dt = 2\int_{\sqrt2}^{2}(t^4 - 2t^2)\,dt$.
3. Integrate: $2\left[\frac{t^5}{5} - \frac{2t^3}{3}\right]_{\sqrt2}^{2} = 2\left[\left(\frac{32}{5} - \frac{16}{3}\right) - \left(\frac{4\sqrt2}{5} - \frac{4\sqrt2}{3}\right)\right]$.
4. Simplify: $2\left[\frac{16}{15} + \frac{8\sqrt2}{15}\right] = \frac{32 + 16\sqrt2}{15}$.

Final Answer: $\dfrac{16\left(2 + \sqrt 2\right)}{15}$.

Example 8: The $e^x[f + f']$ pattern in a definite integral

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

Solution:

1. Substitute $t = 2x$ (so $x = \frac t2$, $dx = \frac{dt}{2}$); limits $1 \to 2$, $2 \to 4$: the integral becomes $\int_2^4\left(\frac{2}{t} - \frac{2}{t^2}\right)e^t\,\frac{dt}{2} = \int_2^4 e^t\left(\frac{1}{t} - \frac{1}{t^2}\right)dt$.
2. Recognise $f + f'$: $f = \frac1t$, $f' = -\frac{1}{t^2}$, so the anti-derivative is $e^t\cdot\frac1t$.
3. Evaluate: $\left[\frac{e^t}{t}\right]_2^4 = \frac{e^4}{4} - \frac{e^2}{2}$.

Final Answer: $\dfrac{e^4}{4} - \dfrac{e^2}{2}$.

Example 9: An identity before the substitution

Evaluate $\displaystyle\int_0^1 \sin^{-1}\left(\frac{2x}{1 + x^2}\right)dx$.

Solution:

1. Use the identity: for $0 \leq x \leq 1$, $\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x$. The integral is $2\int_0^1\tan^{-1}x\,dx$.
2. By parts: $\int\tan^{-1}x\,dx = x\tan^{-1}x - \frac12\log(1 + x^2)$.
3. Evaluate: $2\left[x\tan^{-1}x - \frac12\log(1+x^2)\right]_0^1 = 2\left[\frac{\pi}{4} - \frac12\log 2\right]$.

Final Answer: $\dfrac{\pi}{2} - \log 2$.

Takeaway: Substitution isn't always the first move — sometimes a trigonometric identity ($\frac{2x}{1+x^2}$ is $\sin$ of a double angle in disguise) simplifies the integrand before any calculus happens.