Solved Examples

How to Use This Section

Thirty fully worked problems spanning the entire chapter, arranged easy → medium → hard. The first two batches drill the standard integrals and substitution; the middle batches cover partial fractions and by parts; the final batches take on NCERT's miscellaneous examples and definite integrals with properties.

Method discipline for every problem:

1. Classify first — standard form? function-derivative pair (substitution)? rational function (partial fractions)? product (by parts)? symmetric definite integral (properties)?
2. Write the plan in one line before computing.
3. Check by differentiating the answer — thirty seconds that catches most sign and factor errors.

Key Point: Examiners reuse structures, not numbers. If you can classify each of the thirty problems below at sight, you can classify anything the board paper throws at you.

Batch 1 — Standard Forms and Direct Integration (Easy)

Example 1: Term-by-term basics

Find $\displaystyle\int \left(2x + 3\cos x + e^x\right)dx$.

Solution:

1. Integrate term by term: $\int 2x\,dx = x^2$; $\int 3\cos x\,dx = 3\sin x$; $\int e^x dx = e^x$.
2. Combine with one constant: a single $C$ covers all three.

Final Answer: $x^2 + 3\sin x + e^x + C$.

Example 2: Trig identity before integrating

Find $\displaystyle\int \tan^2 x\,dx$.

Solution:

1. Convert: $\tan^2 x = \sec^2 x - 1$ (no standard integral of $\tan^2 x$ exists directly).
2. Integrate: $\int(\sec^2 x - 1)dx = \tan x - x$.

Final Answer: $\tan x - x + C$.

Example 3: Power reduction

Find $\displaystyle\int \cos^2 x\,dx$.

Solution:

1. Reduce the power: $\cos^2 x = \frac{1 + \cos 2x}{2}$.
2. Integrate: $\frac{x}{2} + \frac{\sin 2x}{4}$.

Final Answer: $\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$.

Example 4: Completing the square (7.4 form)

Find $\displaystyle\int \frac{dx}{x^2 + 4x + 8}$.

Solution:

1. Complete the square: $x^2 + 4x + 8 = (x + 2)^2 + 4$.
2. Match the standard form $\int\frac{dt}{t^2 + a^2} = \frac1a\tan^{-1}\frac ta$ with $t = x + 2$, $a = 2$.

Final Answer: $\dfrac{1}{2}\tan^{-1}\left(\dfrac{x + 2}{2}\right) + C$.

Example 5: The odd-power trick

Find $\displaystyle\int \sin^3 x\,dx$.

Solution:

1. Peel one factor: $\sin^3 x = (1 - \cos^2 x)\sin x$.
2. Substitute $t = \cos x$, $dt = -\sin x\,dx$: $\int(t^2 - 1)\,dt = \frac{t^3}{3} - t$.
3. Return: $\frac{\cos^3 x}{3} - \cos x$.

Final Answer: $-\cos x + \dfrac{\cos^3 x}{3} + C$.

Example 6: Product to sum

Find $\displaystyle\int \sin 3x\cos 4x\,dx$.

Solution:

1. Convert the product: $\sin 3x\cos 4x = \frac{1}{2}[\sin 7x + \sin(-x)] = \frac{1}{2}(\sin 7x - \sin x)$.
2. Integrate: $\frac12\left[-\frac{\cos 7x}{7} + \cos x\right]$.

Final Answer: $\dfrac{\cos x}{2} - \dfrac{\cos 7x}{14} + C$.

Batch 2 — Substitution at Full Power (Easy-Medium)

Example 7: Root of a linear-in-disguise

Find $\displaystyle\int \cos 6x\sqrt{1 + \sin 6x}\,dx$.

Solution:

1. Substitute: $t = 1 + \sin 6x$, $dt = 6\cos 6x\,dx$.
2. Integrate: $\frac16\int t^{1/2}dt = \frac16\cdot\frac23 t^{3/2} = \frac19 t^{3/2}$.
3. Return: $t = 1 + \sin 6x$.

Final Answer: $\dfrac{1}{9}\left(1 + \sin 6x\right)^{3/2} + C$.

Example 8: Square roots in the denominator

Find $\displaystyle\int \frac{dx}{\sqrt x\left(1 + \sqrt x\right)}$.

Solution:

1. Substitute: $t = 1 + \sqrt x$, $dt = \frac{dx}{2\sqrt x}$.
2. Integrate: $2\int\frac{dt}{t} = 2\log|t|$.

Final Answer: $2\log\left|1 + \sqrt x\right| + C$.

Example 9: Exponential of an inverse

Find $\displaystyle\int \frac{e^{\tan^{-1}x}}{1 + x^2}\,dx$.

Solution:

1. Spot the pair: $\frac{1}{1+x^2}$ is the derivative of $\tan^{-1}x$.
2. Substitute $t = \tan^{-1}x$: $\int e^t dt = e^t$.

Final Answer: $e^{\tan^{-1}x} + C$.

Example 10: A clever power split

Find $\displaystyle\int \frac{\left(x^4 - x\right)^{1/4}}{x^5}\,dx$.

Solution:

1. Extract $x$ from the bracket: $(x^4 - x)^{1/4} = x\left(1 - \frac{1}{x^3}\right)^{1/4}$, so the integrand is $\frac{1}{x^4}\left(1 - \frac{1}{x^3}\right)^{1/4} = \left(1 - x^{-3}\right)^{1/4}\cdot\frac{1}{x^4}$.
2. Substitute: $t = 1 - \frac{1}{x^3}$, $dt = \frac{3}{x^4}dx$.
3. Integrate: $\frac13\int t^{1/4}dt = \frac13\cdot\frac45 t^{5/4} = \frac{4}{15}t^{5/4}$.

Final Answer: $\dfrac{4}{15}\left(1 - \dfrac{1}{x^3}\right)^{5/4} + C$.

Takeaway: When a bracket contains mixed powers, factor out the highest power of $x$ — the leftover $\frac{1}{x^n}$ frequently is the derivative needed for a clean substitution.

Example 11: Numerator = derivative of denominator

Find $\displaystyle\int \frac{x^2}{1 + x^3}\,dx$.

Solution:

1. Spot the pair: $\frac{d}{dx}(1 + x^3) = 3x^2$.
2. Substitute $t = 1 + x^3$: $\frac13\int\frac{dt}{t} = \frac13\log|t|$.

Final Answer: $\dfrac{1}{3}\log\left|1 + x^3\right| + C$.

Example 12: A 7.4 form under a root

Find $\displaystyle\int \frac{dx}{\sqrt{9 + 8x - x^2}}$.

Solution:

1. Complete the square: $9 + 8x - x^2 = 25 - (x - 4)^2$.
2. Match $\int\frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac ta$ with $t = x - 4$, $a = 5$.

Final Answer: $\sin^{-1}\left(\dfrac{x - 4}{5}\right) + C$.

Batch 3 — Partial Fractions and By Parts (Medium)

Example 13: Three distinct linear factors

Find $\displaystyle\int \frac{3x - 1}{(x - 1)(x - 2)(x - 3)}\,dx$.

Solution:

1. Set up: $\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.
2. Cover-up at each root: $A = \frac{3(1)-1}{(1-2)(1-3)} = \frac{2}{2} = 1$; $B = \frac{5}{(1)(-1)} = -5$; $C = \frac{8}{(2)(1)} = 4$.
3. Integrate: $\log|x-1| - 5\log|x-2| + 4\log|x-3|$.

Final Answer: $\log|x - 1| - 5\log|x - 2| + 4\log|x - 3| + C$.

Example 14: Improper fraction with a quadratic factor

Find $\displaystyle\int \frac{x^4}{(x - 1)(x^2 + 1)}\,dx$.

Solution:

1. Divide first (degree 4 over degree 3): $\frac{x^4}{(x-1)(x^2+1)} = (x + 1) + \frac{1}{(x-1)(x^2+1)}$.
2. Decompose the remainder: $\frac{1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+1}$; solving $1 = A(x^2+1) + (Bx+C)(x-1)$ gives $A = \frac12$, $B = C = -\frac12$.
3. Integrate each piece: $\int(x+1)dx = \frac{x^2}{2} + x$; $\frac12\log|x-1|$; $-\frac12\int\frac{x\,dx}{x^2+1} = -\frac14\log(x^2+1)$; $-\frac12\tan^{-1}x$.

Final Answer: $\dfrac{x^2}{2} + x + \dfrac{1}{2}\log|x - 1| - \dfrac{1}{4}\log\left(x^2 + 1\right) - \dfrac{1}{2}\tan^{-1}x + C$.

Example 15: By parts twice with a polynomial

Find $\displaystyle\int x^2\sin x\,dx$.

Solution:

1. Round 1 ($f = x^2$): $-x^2\cos x + 2\int x\cos x\,dx$.
2. Round 2 ($f = x$): $\int x\cos x\,dx = x\sin x + \cos x$.
3. Combine: $-x^2\cos x + 2x\sin x + 2\cos x$.

Final Answer: $\left(2 - x^2\right)\cos x + 2x\sin x + C$.

Example 16: Nested logs

Find $\displaystyle\int \left[\log(\log x) + \frac{1}{(\log x)^2}\right]dx$.

Solution:

1. Split: $I = \int\log(\log x)\,dx + \int\frac{dx}{(\log x)^2}$.
2. By parts on the first (with $1$ as second function): $x\log(\log x) - \int\frac{x}{x\log x}dx = x\log(\log x) - \int\frac{dx}{\log x}$.
3. By parts again on $\int\frac{dx}{\log x}$: it equals $\frac{x}{\log x} + \int\frac{dx}{(\log x)^2}$.
4. Combine: the two $\int\frac{dx}{(\log x)^2}$ terms cancel: $I = x\log(\log x) - \frac{x}{\log x}$.

Final Answer: $x\log(\log x) - \dfrac{x}{\log x} + C$.

Takeaway: Neither piece alone has an elementary answer — but their sum does. The pairing is deliberate: integrate by parts and watch the impossible parts cancel.

Example 17: Exponential-times-trig with a coefficient

Find $\displaystyle\int e^{2x}\sin x\,dx$.

Solution:

1. Round 1 ($f = e^{2x}$): $I = -e^{2x}\cos x + 2\int e^{2x}\cos x\,dx$.
2. Round 2: $\int e^{2x}\cos x\,dx = e^{2x}\sin x - 2I$.
3. Solve the loop: $I = -e^{2x}\cos x + 2e^{2x}\sin x - 4I$, so $5I = e^{2x}(2\sin x - \cos x)$.

Final Answer: $\dfrac{e^{2x}}{5}\left(2\sin x - \cos x\right) + C$.

Example 18: The quotient-shaped $e^x[f + f']$

Find $\displaystyle\int \frac{(x - 3)e^x}{(x - 1)^3}\,dx$.

Solution:

1. Split the numerator: $\frac{x - 3}{(x-1)^3} = \frac{(x - 1) - 2}{(x-1)^3} = \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}$.
2. Test: with $f = \frac{1}{(x-1)^2}$, $f' = -\frac{2}{(x-1)^3}$ — the bracket is $f + f'$.
3. Quote the pattern: $\int e^x[f + f']dx = e^x f$.

Final Answer: $\dfrac{e^x}{(x - 1)^2} + C$.

Batch 4 — The Chapter's Hardest Indefinite Integrals (Hard)

Example 19: The cot-plus-tan classic

Find $\displaystyle\int \left(\sqrt{\cot x} + \sqrt{\tan x}\right)dx$.

Solution:

1. Factor: $I = \int\sqrt{\tan x}\left(1 + \cot x\right)dx$.
2. Substitute: $\tan x = t^2$, so $\sec^2 x\,dx = 2t\,dt$, i.e. $dx = \frac{2t\,dt}{1 + t^4}$: $I = \int t\left(1 + \frac{1}{t^2}\right)\frac{2t\,dt}{1+t^4} = 2\int\frac{t^2 + 1}{t^4 + 1}dt$.
3. Divide by $t^2$: $2\int\frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}}dt = 2\int\frac{\left(1 + \frac{1}{t^2}\right)dt}{\left(t - \frac1t\right)^2 + 2}$.
4. Substitute $y = t - \frac1t$ (then $dy = \left(1 + \frac{1}{t^2}\right)dt$): $2\int\frac{dy}{y^2 + 2} = \frac{2}{\sqrt2}\tan^{-1}\frac{y}{\sqrt2}$.

Final Answer: $\sqrt 2\,\tan^{-1}\left(\dfrac{\tan x - 1}{\sqrt{2\tan x}}\right) + C$.

Takeaway: The pattern $\int\frac{t^2 + 1}{t^4 + 1}dt$ — divide by $t^2$, then substitute $y = t - \frac1t$ — is a JEE evergreen. The twin $\int\frac{t^2 - 1}{t^4 + 1}dt$ uses $y = t + \frac1t$ instead.

Example 20: A $\sin^{-1}$ ending

Find $\displaystyle\int \frac{\sin 2x\cos 2x}{\sqrt{9 - \cos^4 2x}}\,dx$.

Solution:

1. Substitute: $t = \cos^2 2x$, so $dt = -4\cos 2x\sin 2x\,dx$, i.e. $\sin 2x\cos 2x\,dx = -\frac{dt}{4}$.
2. Integrate: $-\frac14\int\frac{dt}{\sqrt{9 - t^2}} = -\frac14\sin^{-1}\frac t3$.

Final Answer: $-\dfrac{1}{4}\sin^{-1}\left(\dfrac{\cos^2 2x}{3}\right) + C$.

Example 21: Both log endings from one decomposition

Find $\displaystyle\int \frac{x}{(x^2 + 1)(x^2 + 2)}\,dx$.

Solution:

1. Substitute first: $y = x^2$, $dy = 2x\,dx$: $\frac12\int\frac{dy}{(y+1)(y+2)}$.
2. Decompose: $\frac{1}{(y+1)(y+2)} = \frac{1}{y+1} - \frac{1}{y+2}$.
3. Integrate and return: $\frac12\log\left|\frac{y+1}{y+2}\right| = \frac12\log\frac{x^2+1}{x^2+2}$.

Final Answer: $\dfrac{1}{2}\log\left(\dfrac{x^2 + 1}{x^2 + 2}\right) + C$.

Example 22: The $x(x^n + 1)$ family

Find $\displaystyle\int \frac{dx}{x\left(x^3 + 1\right)}$.

Solution:

1. Multiply and divide by $x^2$: $I = \int\frac{x^2\,dx}{x^3(x^3+1)}$.
2. Substitute $t = x^3$, $dt = 3x^2dx$: $\frac13\int\frac{dt}{t(t+1)}$.
3. Decompose and integrate: $\frac13\int\left(\frac1t - \frac{1}{t+1}\right)dt = \frac13\log\left|\frac{t}{t+1}\right|$.

Final Answer: $\dfrac{1}{3}\log\left|\dfrac{x^3}{x^3 + 1}\right| + C$.

Takeaway: For $\int\frac{dx}{x(x^n + 1)}$, multiplying by $\frac{x^{n-1}}{x^{n-1}}$ converts the problem to a two-term partial fraction in $t = x^n$ — one substitution replaces an $n$-factor decomposition.

Example 23: The loop-back with secant

Find $\displaystyle\int \sec^3 x\,dx$.

Solution:

1. Split: $I = \int\sec x\cdot\sec^2 x\,dx$; by parts with $f = \sec x$, $g = \sec^2 x$: $I = \sec x\tan x - \int\sec x\tan^2 x\,dx$.
2. Convert: $\tan^2 x = \sec^2 x - 1$: $I = \sec x\tan x - I + \int\sec x\,dx$.
3. Solve the loop: $2I = \sec x\tan x + \log|\sec x + \tan x|$.

Final Answer: $\dfrac{1}{2}\left[\sec x\tan x + \log\left|\sec x + \tan x\right|\right] + C$.

Example 24: Splitting a linear numerator over a root

Find $\displaystyle\int \frac{x + 2}{\sqrt{x^2 + 4x + 5}}\,dx$.

Solution:

1. Spot the derivative: $\frac{d}{dx}(x^2 + 4x + 5) = 2x + 4 = 2(x + 2)$ — the numerator is exactly half of it.
2. Substitute $t = x^2 + 4x + 5$: $\frac12\int\frac{dt}{\sqrt t} = \sqrt t$.

Final Answer: $\sqrt{x^2 + 4x + 5} + C$.

Batch 5 — Definite Integrals with Properties (Hard)

Example 25: Modulus with a trig zero

Evaluate $\displaystyle\int_{-1}^{3/2} \left|x\sin\pi x\right|dx$.

Solution:

1. Sign chart: $x\sin\pi x \geq 0$ on $[-1, 1]$ (both factors change sign together) and $\leq 0$ on $\left[1, \frac32\right]$.
2. Split (P2): $I = \int_{-1}^{1} x\sin\pi x\,dx - \int_1^{3/2} x\sin\pi x\,dx$.
3. Anti-derivative by parts: $\int x\sin\pi x\,dx = -\frac{x\cos\pi x}{\pi} + \frac{\sin\pi x}{\pi^2}$.
4. Evaluate: with $F(x) = -\frac{x\cos\pi x}{\pi} + \frac{\sin\pi x}{\pi^2}$: first piece $F(1) - F(-1) = \frac1\pi - \left(-\frac1\pi\right) = \frac{2}{\pi}$; second piece $F\left(\frac32\right) - F(1) = -\frac{1}{\pi^2} - \frac{1}{\pi}$, entering with a minus sign. Total: $\frac{2}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2}$.

Final Answer: $\dfrac{3}{\pi} + \dfrac{1}{\pi^2}$.

Example 26: The $a$-$b$ symmetric denominator

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

Solution:

1. King's rule (P4): the trig block is unchanged under $x \mapsto \pi - x$, so adding the mirrored copy: $2I = \pi\int_0^{\pi}\frac{dx}{a^2\cos^2x + b^2\sin^2x}$.
2. Use P6 ($f(\pi - x) = f(x)$): $2I = 2\pi\int_0^{\pi/2}\frac{dx}{a^2\cos^2x + b^2\sin^2x}$.
3. Divide by $\cos^2 x$ and substitute $t = \tan x$: $I = \pi\int_0^{\infty}\frac{dt}{a^2 + b^2t^2} = \frac{\pi}{ab}\left[\tan^{-1}\frac{bt}{a}\right]_0^{\infty} = \frac{\pi}{ab}\cdot\frac{\pi}{2}$.

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

Takeaway: An alternative route splits $\left[0, \frac{\pi}{2}\right]$ at $\frac{\pi}{4}$ and uses $\tan$ and $\cot$ substitutions to avoid the improper limit — same value, more steps. Either route needs the King's rule first to kill the naked $x$.

Example 27: Odd about the centre (P6)

Evaluate $\displaystyle\int_0^{2\pi} \cos^5 x\,dx$.

Solution:

1. Check the $[0, 2a]$ symmetry with $a = \pi$: $\cos^5(2\pi - x) = \cos^5 x$, so P6 gives $I = 2\int_0^{\pi}\cos^5 x\,dx$.
2. Check again with $a = \frac{\pi}{2}$: $\cos^5(\pi - x) = -\cos^5 x$ — the anti-symmetric case of P6.
3. Conclude: $\int_0^{\pi}\cos^5 x\,dx = 0$, hence $I = 0$.

Final Answer: $0$.

Example 28: A log-one-liner

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

Solution:

1. Spot the pair: numerator $= -\frac{d}{dx}(1 + \cos x)$.
2. Integrate: $-\log(1 + \cos x)$; evaluate from $0$ to $\frac{\pi}{2}$: $-\log 1 + \log 2$.

Final Answer: $\log 2$.

Example 29: Rationalise, then integrate (Miscellaneous style)

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

Solution:

1. Rationalise: multiply by $\frac{\sqrt{1+x} - \sqrt x}{\sqrt{1+x} - \sqrt x}$: denominator becomes $(1 + x) - x = 1$, so the integrand is $\sqrt{1+x} - \sqrt x$.
2. Integrate: $\frac23(1+x)^{3/2} - \frac23 x^{3/2}$.
3. Evaluate: $\frac23\left[2\sqrt2 - 1\right] - \frac23\left[1 - 0\right] = \frac{4\sqrt2 - 4}{3}$.

Final Answer: $\dfrac{4\left(\sqrt 2 - 1\right)}{3}$.

Example 30: The half-angle log integral

Evaluate $\displaystyle\int_0^{\pi} \log(1 + \cos x)\,dx$.

Solution:

1. Half-angle: $1 + \cos x = 2\cos^2\frac x2$, so $I = \pi\log 2 + 2\int_0^{\pi}\log\cos\frac x2\,dx$.
2. Substitute $t = \frac x2$: $2\int_0^{\pi}\log\cos\frac x2 dx = 4\int_0^{\pi/2}\log\cos t\,dt$.
3. Quote the famous result (equal to the log-sine integral by P4): $\int_0^{\pi/2}\log\cos t\,dt = -\frac{\pi}{2}\log 2$.
4. Combine: $I = \pi\log2 + 4\left(-\frac{\pi}{2}\log2\right) = \pi\log 2 - 2\pi\log 2$.

Final Answer: $-\pi\log 2$.

Takeaway: The value $\int_0^{\pi/2}\log\sin x\,dx = \int_0^{\pi/2}\log\cos x\,dx = -\frac{\pi}{2}\log2$ is a building block — several miscellaneous-level definite integrals reduce to it after a half-angle or double-angle step.