JEE Corner — Integrals for JEE Main and Advanced

Where Integrals Sits in JEE

Integral Calculus is a guaranteed-marks unit in JEE Main: expect 2-3 questions every session across indefinite integration, definite integration and (in the next chapter) area under curves. The official syllabus covers integration as the inverse of differentiation, all the standard techniques — substitution, partial fractions, by parts — the standard-form integrals of Sections 7.4 and 7.6.2, the Fundamental Theorem of Calculus, and properties of definite integrals. JEE Advanced adds heavier compositions of the same tools and multi-concept problems mixing integrals with functions, limits and differential equations.

The JEE add-ons beyond NCERT's toolkit

1. Weierstrass substitution. For $\int\frac{dx}{a + b\cos x}$ or $\int\frac{dx}{a + b\sin x}$: put $t = \tan\frac x2$, so $\cos x = \frac{1 - t^2}{1 + t^2}$, $\sin x = \frac{2t}{1 + t^2}$, $dx = \frac{2\,dt}{1 + t^2}$. The integral becomes rational in $t$ and lands in a Section 7.4 standard form.
2. The $x^4 + 1$ family. For $\int\frac{x^2 + 1}{x^4 + 1}dx$: divide top and bottom by $x^2$, then substitute $y = x - \frac1x$ (denominator becomes $y^2 + 2$). For $\int\frac{x^2 - 1}{x^4 + 1}dx$: same but $y = x + \frac1x$ (denominator $y^2 - 2$). Which $y$? The one whose $dy$ matches the numerator.
3. The $\frac{1}{x(x^n + 1)}$ shortcut. Multiply by $\frac{x^{n-1}}{x^{n-1}}$, substitute $t = x^n$: answer $\frac1n\log\left|\frac{x^n}{x^n + 1}\right| + C$ in one line.
4. Trig-ratio integrals $\int\frac{p\cos x + q\sin x}{a\cos x + b\sin x}dx$: write the numerator as $A(\text{denominator}) + B(\text{derivative of denominator})$; answer $Ax + B\log|a\cos x + b\sin x| + C$.
5. Periodicity. If $f$ has period $T$: $\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx$. Combined with $\int_0^{\pi}|\sin x|dx = 2$: $\int_0^{n\pi}|\sin x|\,dx = 2n$ instantly.

Key Point: JEE rewards classification speed. Every integral in the paper is one of perhaps fifteen patterns; the exam tests whether you can name the pattern in five seconds and execute in sixty.

[JEE Tip] Options are your ally: differentiating the four options is often faster than integrating the question — especially for $e^x[f + f']$ shapes and root-formula answers.

Definite-Integral Weapons and Classic Traps

The high-frequency toolkit

1. King's rule reflex. $\int_0^a f(x)dx = \int_0^a f(a-x)dx$, and its most-used corollary: $\int_0^{\pi} x\,f(\sin x)\,dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x)\,dx$ — any time $x$ multiplies a function of $\sin x$ on $[0, \pi]$, the $x$ is deletable.
2. Self-complementary ratios. $\int_0^{\pi/2}\frac{\sin^n x}{\sin^n x + \cos^n x}dx = \frac{\pi}{4}$ for every $n$ (even irrational $n$) — add the mirror copy. Same for $\frac{1}{1 + \tan^n x}$ and $\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}$ shapes.
3. Log pairs: $\int_0^{\pi/2}\log\tan x\,dx = 0$; $\int_0^{\pi/2}\log\sin x\,dx = -\frac{\pi}{2}\log 2$; $\int_0^{\pi/4}\log(1 + \tan x)dx = \frac{\pi}{8}\log 2$ — all three are one King's-rule application each; JEE quotes them as sub-steps of harder problems.
4. Leibniz differentiation of variable limits: $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v(x))\,v'(x) - f(u(x))\,u'(x)$. With $u = 0$, $v = x$ this is the First FTC; JEE loves $v(x) = x^2$ or an equation involving $\int_0^x f$ that must be differentiated to find $f$.
5. Estimation without evaluation. On $[0, 1]$: $x^3 \leq x^2$, so $\int_0^1 g\left(x^3\right)dx \geq$ or $\leq \int_0^1 g\left(x^2\right)dx$ according as $g$ is decreasing or increasing. Comparison questions need monotonicity, not computation.

Traps that cost real marks

1. $\sin^2 x$ is even, $\sin^3 x$ is odd — a square never inherits oddness. Check the whole integrand, not the innermost function.
2. Odd function needs a symmetric interval. $\int_{-1}^{2} x^3\,dx \neq 0$.
3. Discontinuity inside the interval invalidates the FTC — watch $\tan x$ on intervals containing $\frac{\pi}{2}$, and $\frac{1}{x}$ across $0$.
4. Modulus before formula. $\int_{-1}^{2}\left|x^3 - x\right|dx$ must be split at the sign changes; slapping $F(b) - F(a)$ on the unsplit integrand gives a wrong (and offered!) option.

[JEE Tip] In the +4/−1 economy, a symmetry observation that produces $0$ or $\frac{\pi}{4}$ without computation is the best-paid move in the paper. Scan for symmetry before reaching for techniques.

JEE-Pattern Worked Examples

Example 1: Weierstrass substitution [JEE Main pattern]

Evaluate $\displaystyle\int \frac{dx}{3 + 2\cos x}$.

Solution:

1. Substitute $t = \tan\frac x2$: $\cos x = \frac{1 - t^2}{1 + t^2}$, $dx = \frac{2\,dt}{1 + t^2}$.
2. Simplify: denominator $3 + 2\cdot\frac{1-t^2}{1+t^2} = \frac{3(1 + t^2) + 2(1 - t^2)}{1+t^2} = \frac{t^2 + 5}{1 + t^2}$, so $I = \int\frac{2\,dt}{t^2 + 5}$.
3. Standard form: $\frac{2}{\sqrt5}\tan^{-1}\frac{t}{\sqrt5}$.

Final Answer: $\dfrac{2}{\sqrt 5}\tan^{-1}\left(\dfrac{\tan\frac{x}{2}}{\sqrt 5}\right) + C$.

Example 2: The $x^4 + 1$ denominator [JEE Main pattern]

Evaluate $\displaystyle\int \frac{x^2 + 1}{x^4 + 1}\,dx$.

Solution:

1. Divide by $x^2$: $I = \int\frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}}\,dx$.
2. Substitute $y = x - \frac1x$ (its differential $dy = \left(1 + \frac{1}{x^2}\right)dx$ is exactly the numerator; and $x^2 + \frac{1}{x^2} = y^2 + 2$): $I = \int\frac{dy}{y^2 + 2}$.
3. Standard form: $\frac{1}{\sqrt2}\tan^{-1}\frac{y}{\sqrt2}$.

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

Example 3: Periodic modulus [JEE Main pattern]

Evaluate $\displaystyle\int_0^{2\pi} |\sin x|\,dx$.

Solution:

1. Period observation: $|\sin x|$ has period $\pi$, and $\int_0^{\pi}|\sin x|dx = \int_0^{\pi}\sin x\,dx = 2$.
2. Apply periodicity: $\int_0^{2\pi} = 2\int_0^{\pi} = 2\cdot2$.

Final Answer: $4$.

Example 4: Leibniz rule with a variable limit [JEE Main pattern]

If $F(x) = \displaystyle\int_0^{x^2} \cos t\,dt$, find $F'(x)$.

Solution:

1. Leibniz rule: $\frac{d}{dx}\int_0^{v(x)} f(t)dt = f(v(x))\cdot v'(x)$ with $v(x) = x^2$.
2. Apply: $F'(x) = \cos\left(x^2\right)\cdot 2x$.

Final Answer: $F'(x) = 2x\cos x^2$. (Check: $F(x) = \sin x^2$ directly, and $\frac{d}{dx}\sin x^2 = 2x\cos x^2$. ✓)

Example 5: Even extension with modulus [JEE Main pattern]

Evaluate $\displaystyle\int_{-\pi/2}^{\pi/2} \left(\sin|x| + \cos|x|\right)dx$.

Solution:

1. Symmetry: both $\sin|x|$ and $\cos|x|$ are even functions of $x$, so $I = 2\int_0^{\pi/2}(\sin x + \cos x)\,dx$ (on $[0, \frac{\pi}{2}]$, $|x| = x$).
2. Evaluate: $2\left[-\cos x + \sin x\right]_0^{\pi/2} = 2\left[(0 + 1) - (-1 + 0)\right] = 2\cdot2$.

Final Answer: $4$.

Example 6: The one-line $x(x^n + 1)$ integral [JEE Main pattern]

Evaluate $\displaystyle\int \frac{dx}{x\left(x^6 + 1\right)}$.

Solution:

1. Multiply and divide by $x^5$: $I = \int\frac{x^5\,dx}{x^6\left(x^6 + 1\right)}$.
2. Substitute $t = x^6$: $\frac16\int\frac{dt}{t(t + 1)} = \frac16\log\left|\frac{t}{t+1}\right|$.

Final Answer: $\dfrac{1}{6}\log\left|\dfrac{x^6}{x^6 + 1}\right| + C$.

Example 7: Power-independent ratio [JEE Main pattern]

Evaluate $\displaystyle\int_0^{\pi/2} \frac{dx}{1 + \tan^{\sqrt 2} x}$.

Solution:

1. Don't integrate — reflect: write the integrand as $\frac{\cos^{\sqrt2}x}{\cos^{\sqrt2}x + \sin^{\sqrt2}x}$; the King's rule swaps $\sin \leftrightarrow \cos$.
2. Add the mirror copy: $2I = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}$.

Final Answer: $\dfrac{\pi}{4}$ — the irrational exponent $\sqrt2$ is a decoy; the argument never uses it.

Example 8: Differentiating an integral equation [JEE Advanced pattern]

If $\displaystyle\int_0^x f(t)\,dt = x^2 + \int_x^1 t\,f(t)\,dt$, find $f(1)$.

Solution:

1. Differentiate both sides (Leibniz: the second integral has $x$ as its lower limit, so its derivative is $-x f(x)$): $f(x) = 2x - x\,f(x)$.
2. Solve for $f$: $f(x)\left(1 + x\right) = 2x$, so $f(x) = \frac{2x}{1 + x}$.
3. Evaluate: $f(1) = \frac{2}{2} = 1$.

Final Answer: $f(1) = 1$.

Takeaway: When an unknown function sits inside an integral equation, differentiate the whole equation — the FTC converts it into an algebraic (or differential) equation for $f$.