Easy Practice — Set 1

Quick Recap — Indefinite Integrals

• $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ ($n\ne-1$); $\int\tfrac1x\,dx=\ln|x|+C$.
• $\int e^x\,dx=e^x+C$; $\int a^x\,dx=\dfrac{a^x}{\ln a}+C$.
• $\int\cos x\,dx=\sin x+C$, $\int\sin x\,dx=-\cos x+C$, $\int\sec^2x\,dx=\tan x+C$.
• $\int\tfrac{dx}{1+x^2}=\tan^{-1}x+C$, $\int\tfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C$. Integration reverses differentiation.

Beyond-NCERT JEE Formulae

Five results that collapse multi-step JEE integrals into one-liners. NCERT rarely states them, yet JEE Main leans on them every year.

1. The $e^{x}[f(x)+f'(x)]$ rule

$$\int e^{x}\big[f(x)+f'(x)\big]\,dx=e^{x}f(x)+C$$

Because $\dfrac{d}{dx}\big[e^{ax}f(x)\big]=e^{ax}\big[a\,f(x)+f'(x)\big]$, the scaled version also holds:

$$\int e^{ax}\big[a\,f(x)+f'(x)\big]\,dx=e^{ax}f(x)+C.$$

[JEE Tip] When the integrand is $e^{x}$ times a sum, check whether one part is the derivative of the other. Name $f$ so that $f'$ sits right beside it, and the answer is simply $e^{x}f(x)$ with no integration required. Disguised forms such as $\dfrac{1+\sin x}{1+\cos x}$ hide an $f+f'$ pair that surfaces after a half-angle rewrite.

2. Wallis' formula

For powers on the first quadrant, where $\displaystyle\int_{0}^{\pi/2}\sin^{n}x\,dx=\int_{0}^{\pi/2}\cos^{n}x\,dx$:

$$\int_{0}^{\pi/2}\sin^{n}x\,dx=\dfrac{(n-1)(n-3)(n-5)\cdots}{n(n-2)(n-4)\cdots}\times\begin{cases}\dfrac{\pi}{2}, & n\text{ even}\\[4pt] 1, & n\text{ odd}\end{cases}$$

Each product descends to $2$ for even $n$ and to $1$ for odd $n$. The $\dfrac{\pi}{2}$ tag appears only when $n$ is even; odd powers give a pure rational number.

3. The split at $\displaystyle\int_{0}^{2a}$

$$\int_{0}^{2a}f(x)\,dx=\begin{cases}2\displaystyle\int_{0}^{a}f(x)\,dx, & f(2a-x)=f(x)\\[4pt] 0, & f(2a-x)=-f(x)\end{cases}$$

Test the integrand's symmetry about the midpoint $x=a$: a mirror-symmetric graph doubles the half-integral, while an anti-symmetric one cancels to zero.

4. Integrals of periodic functions

If $f$ has period $T$, meaning $f(x+T)=f(x)$, then over any whole number of periods

$$\int_{0}^{nT}f(x)\,dx=n\int_{0}^{T}f(x)\,dx,\qquad n=1,2,3,\dots$$

and more generally $\displaystyle\int_{a}^{a+nT}f(x)\,dx=n\int_{0}^{T}f(x)\,dx$ for every real $a$. Always use the true period: both $|\sin x|$ and $\sin^{2}x$ repeat with period $\pi$, not $2\pi$.

5. Weierstrass (half-angle) substitution

Putting $t=\tan\dfrac{x}{2}$ rationalises any integral built from $\sin x$ and $\cos x$:

$$\sin x=\dfrac{2t}{1+t^{2}},\qquad \cos x=\dfrac{1-t^{2}}{1+t^{2}},\qquad dx=\dfrac{2\,dt}{1+t^{2}}.$$

This turns $\displaystyle\int\dfrac{dx}{a+b\cos x}$ (and likewise the $a+b\sin x$ form) into a rational integral in $t$. For a definite integral with $x:0\to\pi$, the new limits become $t:0\to\infty$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Spotting $f+f'$. Evaluate $\displaystyle\int e^{x}\left(\tan^{-1}x+\dfrac{1}{1+x^{2}}\right)dx$.

Let $f(x)=\tan^{-1}x$. Then $f'(x)=\dfrac{1}{1+x^{2}}$, so the bracket is exactly $f(x)+f'(x)$. Applying $\displaystyle\int e^{x}[f+f']\,dx=e^{x}f$,

$$\int e^{x}\left(\tan^{-1}x+\dfrac{1}{1+x^{2}}\right)dx=e^{x}\tan^{-1}x+C.$$

Example 2 — A disguised $f+f'$. Evaluate $\displaystyle\int e^{x}\,\dfrac{1+\sin x}{1+\cos x}\,dx$.

Rewrite with half-angles, using $1+\cos x=2\cos^{2}\dfrac{x}{2}$ together with $1+\sin x=1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$:

$$\dfrac{1+\sin x}{1+\cos x}=\dfrac{1}{2}\sec^{2}\dfrac{x}{2}+\tan\dfrac{x}{2}.$$

Take $f(x)=\tan\dfrac{x}{2}$, so that $f'(x)=\dfrac{1}{2}\sec^{2}\dfrac{x}{2}$ and the integrand is $e^{x}[f(x)+f'(x)]$. Therefore

$$\int e^{x}\,\dfrac{1+\sin x}{1+\cos x}\,dx=e^{x}\tan\dfrac{x}{2}+C.$$

Example 3 — The scaled $e^{ax}$ version. Evaluate $\displaystyle\int e^{3x}\big(3\sin x+\cos x\big)\,dx$.

Here $a=3$ with $f(x)=\sin x$, so $a\,f(x)+f'(x)=3\sin x+\cos x$ fits the pattern $e^{ax}[a\,f+f']$. Hence

$$\int e^{3x}\big(3\sin x+\cos x\big)\,dx=e^{3x}\sin x+C.$$

A quick check confirms it: $\dfrac{d}{dx}\big(e^{3x}\sin x\big)=3e^{3x}\sin x+e^{3x}\cos x=e^{3x}\big(3\sin x+\cos x\big).$

Example 4 — Wallis for an even power. Evaluate $\displaystyle\int_{0}^{\pi/2}\sin^{6}x\,dx$.

With $n=6$ (even), the descending odd-over-even product carries the $\dfrac{\pi}{2}$ factor:

$$\int_{0}^{\pi/2}\sin^{6}x\,dx=\dfrac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\cdot\dfrac{\pi}{2}=\dfrac{15}{48}\cdot\dfrac{\pi}{2}=\dfrac{5\pi}{32}.$$

Example 5 — Combining the $\displaystyle\int_{0}^{2a}$ split with Wallis. Evaluate $\displaystyle\int_{0}^{\pi}\sin^{3}x\,dx$.

Here $2a=\pi$, so $a=\dfrac{\pi}{2}$. Since $\sin^{3}(\pi-x)=\sin^{3}x$, the integrand is symmetric about $x=\dfrac{\pi}{2}$, giving $\displaystyle\int_{0}^{\pi}\sin^{3}x\,dx=2\int_{0}^{\pi/2}\sin^{3}x\,dx$. Wallis with $n=3$ (odd) gives $\displaystyle\int_{0}^{\pi/2}\sin^{3}x\,dx=\dfrac{2}{3}$, hence

$$\int_{0}^{\pi}\sin^{3}x\,dx=2\cdot\dfrac{2}{3}=\dfrac{4}{3}.$$

Example 6 — The anti-symmetric $\displaystyle\int_{0}^{2a}$ case. Evaluate $\displaystyle\int_{0}^{\pi}\cos^{3}x\,dx$.

Again $2a=\pi$. Now $\cos^{3}(\pi-x)=(-\cos x)^{3}=-\cos^{3}x$, so the integrand is anti-symmetric about $x=\dfrac{\pi}{2}$ and the two halves cancel:

$$\int_{0}^{\pi}\cos^{3}x\,dx=0.$$

Example 7 — A periodic integral. Evaluate $\displaystyle\int_{0}^{n\pi}|\sin x|\,dx$ for a positive integer $n$.

The function $|\sin x|$ has period $T=\pi$ (not $2\pi$), and over one period $\displaystyle\int_{0}^{\pi}|\sin x|\,dx=\int_{0}^{\pi}\sin x\,dx=\big[-\cos x\big]_{0}^{\pi}=2$. Using $\displaystyle\int_{0}^{nT}f=n\int_{0}^{T}f$,

$$\int_{0}^{n\pi}|\sin x|\,dx=n\int_{0}^{\pi}|\sin x|\,dx=2n.$$

For example, $\displaystyle\int_{0}^{5\pi}|\sin x|\,dx=10$.

Example 8 — Weierstrass substitution. Evaluate $\displaystyle\int_{0}^{\pi}\dfrac{dx}{2+\cos x}$.

Put $t=\tan\dfrac{x}{2}$, so $\cos x=\dfrac{1-t^{2}}{1+t^{2}}$ and $dx=\dfrac{2\,dt}{1+t^{2}}$; the range $x:0\to\pi$ corresponds to $t:0\to\infty$. The denominator becomes

$$2+\cos x=\dfrac{2(1+t^{2})+(1-t^{2})}{1+t^{2}}=\dfrac{3+t^{2}}{1+t^{2}},$$

so the integrand collapses to $\dfrac{2\,dt}{3+t^{2}}$. Therefore

$$\int_{0}^{\pi}\dfrac{dx}{2+\cos x}=\int_{0}^{\infty}\dfrac{2\,dt}{3+t^{2}}=\dfrac{2}{\sqrt{3}}\left[\tan^{-1}\dfrac{t}{\sqrt{3}}\right]_{0}^{\infty}=\dfrac{2}{\sqrt{3}}\cdot\dfrac{\pi}{2}=\dfrac{\pi}{\sqrt{3}}.