How do I decide the substitution for R(sin x, cos x) dx?

Test the symmetry: if replacing sin x by -sin x flips the sign, use cos x = t; if cos x by -cos x flips it, use sin x = t; if both replaced together leaves it unchanged, use tan x = t; if none, fall back to t = tan(x/2).

What do I do for the integral of sin^M x cos^N x dx?

If one of M, N is odd, substitute for the even-power term. If both are odd, either substitution works. If both are even, use identities sin squared x equals (1 - cos 2x)/2 and cos squared x equals (1 + cos 2x)/2 to reduce the powers.

What substitution works for the integral of 1/((x-k)^r sqrt(ax^2+bx+c))?

Put x - k = 1/t, so dx = -dt/t^2. This transforms the integral into a form t^(r-1) dt over sqrt(At^2 + Bt + C) for some constants A, B, C - treatable by completing the square.

Integrals of Some Particular Functions

Q: How do I integrate (px+q)/(ax^2+bx+c) dx?

Write px + q as ell times (2ax + b) plus m, where 2ax + b is the derivative of the quadratic. Find ell and m by comparing coefficients. The first part integrates by substitution u = ax^2 + bx + c giving ell ln|u|; the second part is a standard integral handled by completing the square.

Q: What is the Weierstrass substitution and when should I use it?

The Weierstrass or universal substitution is t = tan(x/2), which converts any rational function of sin x and cos x into a rational function of t, integrable by partial fractions. Use it when no obvious symmetry substitution applies. It always works but often produces heavy algebra.

Q: When do I need to complete the square?

Whenever the denominator is a general quadratic ax^2 + bx + c (or its square root) with b nonzero. Completing the square rewrites it in a form matching one of the standard integrals for 1/(u^2 plus or minus a^2) or 1/sqrt(u^2 plus or minus a^2).

MathsIntegralsFor JEE aspirants

Integrals of some particular functions in JEE Maths refers to standard techniques for algebraic quotients like $\frac{p x + q}{a x ^{2} + b x + c}$ and $\frac{p x + q}{a x ^{2} + b x + c}$ , irrational integrals of the form $\frac{1}{( x - k ) ^{r} a x ^{2} + b x + c}$ , and trigonometric integrals $\int R (sin x, cos x) d x$ . The core trick for the algebraic cases is to write $p x + q = ℓ (derivative of quadratic) + m$ , splitting the integral into one that reduces by substitution and one that is a standard form. For trigonometric integrals, the Weierstrass substitution $t = tan (x /2)$ or symmetry-based substitutions ( $cos x = t$ or $sin x = t$ ) apply. These forms appear in almost every JEE problem set.

Key Formulas - Quick Reference

$\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a} + C$
$\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} ln \frac{x - a}{x + a} + C$
$\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a} + C$
$\int \frac{d x}{x ^{2} + a ^{2}} = ln x + x^{2} + a^{2} + C$
$\int \frac{d x}{x ^{2} - a ^{2}} = ln x + x^{2} - a^{2} + C$
For $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ (or with $\cdot$ ): write $p x + q = ℓ (2 a x + b) + m$
Weierstrass: $t = tan (x /2)$ , $sin x = \frac{2 t}{1 + t ^{2}}$ , $cos x = \frac{1 - t ^{2}}{1 + t ^{2}}$ , $d x = \frac{2 d t}{1 + t ^{2}}$

1. Algebraic Integrals with Linear Numerator

The three model forms are:

$\int \frac{p x + q}{a x ^{2} + b x + c} d x, \int \frac{p x + q}{a x ^{2} + b x + c} d x, \int (p x + q) a x^{2} + b x + c d x .$

Technique: write

$p x + q = ℓ \cdot \frac{d}{d x} (a x^{2} + b x + c) + m = ℓ (2 a x + b) + m .$

Compare coefficients of $x$ and the constant term to find $ℓ$ and $m$ . The integral then splits into two parts:

The $ℓ (2 a x + b)$ part uses the substitution $u = a x^{2} + b x + c$ , so $d u = (2 a x + b) d x$ . This gives a power (or logarithm) of $u$ .
The constant part $m$ leaves a standard integral $\int \frac{d x}{a x ^{2} + b x + c}$ or $\int \frac{d x}{a x ^{2} + b x + c}$ , which is completed by "completing the square" and applying one of the standard formulae in the box above.

Ratio of two quadratics

For integrals of the type

$\int \frac{a x ^{2} + b x + c}{p x ^{2} + q x + r} d x or \int \frac{a x ^{2} + b x + c}{p x ^{2} + q x + r} d x,$

write

$a x^{2} + b x + c = M (p x^{2} + q x + r) + N (2 p x + q) + R .$

Find $M, N, R$ by comparing coefficients. The integral splits into three standard pieces.

Solved Example 1

Evaluate

\int \frac{x + 1}{2 x ^{2} + x - 3} d x .

Solution:

Write $x + 1 = A (4 x + 1) + B$ . Comparing coefficients: $4 A = 1$ and $A + B = 1$ , giving $A = \frac{1}{4}, B = \frac{3}{4}$ .

$I = \frac{1}{4} \int \frac{4 x + 1}{2 x ^{2} + x - 3} d x + \frac{3}{4} \int \frac{d x}{2 x ^{2} + x - 3} = I_{1} + I_{2} .$

Part $I_{1}$ : Put $z = 2 x^{2} + x - 3$ , so $d z = (4 x + 1) d x$ :

$I_{1} = \frac{1}{4} \int \frac{d z}{z} = \frac{1}{4} \cdot 2 z = \frac{1}{2} 2 x^{2} + x - 3 .$

Part $I_{2}$ : Complete the square: $2 x^{2} + x - 3 = 2 [(x + \frac{1}{4})^{2} - \frac{25}{16}]$ . Then

$I_{2} = \frac{3}{4 2} \int \frac{d x}{( x + \frac{1}{4} ) ^{2} - ( \frac{5}{4} ) ^{2}} = \frac{3}{4 2} ln x + \frac{1}{4} + (x + \frac{1}{4})^{2} - \frac{25}{16} + C .$

Combining:

$I = \frac{1}{2} 2 x^{2} + x - 3 + \frac{3}{4 2} ln x + \frac{1}{4} + (x + \frac{1}{4})^{2} - \frac{25}{16} + C .$

Solved Example 2

Evaluate

\int \frac{3 x + 2}{4 x ^{2} + 4 x + 5} d x .

Solution:

Write $3 x + 2 = ℓ (8 x + 4) + m$ . Comparing: $8 ℓ = 3$ and $4 ℓ + m = 2$ , giving $ℓ = \frac{3}{8}, m = \frac{1}{2}$ .

$I = \frac{3}{8} \int \frac{8 x + 4}{4 x ^{2} + 4 x + 5} d x + \frac{1}{2} \int \frac{d x}{4 x ^{2} + 4 x + 5} .$

The first is $\frac{3}{8} ln ∣4 x^{2} + 4 x + 5∣$ (by the substitution $u = 4 x^{2} + 4 x + 5$ ).

For the second, complete the square: $4 x^{2} + 4 x + 5 = 4 [(x + \frac{1}{2})^{2} + 1]$ . So

$\frac{1}{2} \int \frac{d x}{4 [ ( x + \frac{1}{2} ) ^{2} + 1 ]} = \frac{1}{8} \int \frac{d x}{( x + \frac{1}{2} ) ^{2} + 1} = \frac{1}{8} tan^{- 1} (x + \frac{1}{2}) .$

$I = \frac{3}{8} ln (4 x^{2} + 4 x + 5) + \frac{1}{8} tan^{- 1} (x + \frac{1}{2}) + C .$

2. Integration of Irrational Algebraic Fractions

Four standard forms with their prescribed substitutions:

Form	Substitution
$\int f (x, (a x + b)^{1/ n}) d x$	$t^{n} = a x + b$
$\int \frac{d x}{( x - k ) ^{r} a x ^{2} + b x + c}$	$x - k = \frac{1}{t}$
$\int \frac{d x}{( A x ^{2} + B ) C x ^{2} + D}$	$x = \frac{1}{t}$ , then $C + D t^{2} = u^{2}$
$\int \frac{( a x ^{2} + b x + c ) d x}{( d x + e ) f x ^{2} + g x + h}$	Write $a x^{2} + b x + c = A_{1} (d x + e) (2 f x + g) + B_{1} (d x + e) + C_{1}$

Solved Example 3

Evaluate

\int \frac{x}{( x - 3 ) x + 1} d x .

Solution:

Put $x + 1 = t^{2}$ , so $x = t^{2} - 1$ , $d x = 2 t d t$ , and $x - 3 = t^{2} - 4$ , $x + 1 = t$ .

$I = \int \frac{( t ^{2} - 1 ) \cdot 2 t d t}{( t ^{2} - 4 ) \cdot t} = 2 \int \frac{t ^{2} - 1}{t ^{2} - 4} d t .$

Divide: $\frac{t ^{2} - 1}{t ^{2} - 4} = 1 + \frac{3}{t ^{2} - 4}$ . So

$I = 2 \int (1 + \frac{3}{t ^{2} - 4}) d t = 2 t + \frac{3}{2} ln \frac{t - 2}{t + 2} + C .$

Back-substitute $t = x + 1$ :

$I = 2 x + 1 + \frac{3}{2} ln \frac{x + 1 - 2}{x + 1 + 2} + C .$

3. Trigonometric Integrals

3.1 Integrals of $R (sin x, cos x)$

If $R (sin x, cos x)$ is a rational function of sine and cosine, the Weierstrass substitution $t = tan (x /2)$ converts it into an integral of a rational function of $t$ . Under this substitution:

$sin x = \frac{2 t}{1 + t ^{2}}, cos x = \frac{1 - t ^{2}}{1 + t ^{2}}, x = 2 tan^{- 1} t, d x = \frac{2 d t}{1 + t ^{2}} .$

Sometimes $t = cot (x /2)$ is more convenient. The universal substitution often produces messy algebra - so use these shortcuts first when a symmetry is available:

If $R (- sin x, cos x) = - R (sin x, cos x)$ : substitute $cos x = t$ .
If $R (sin x, - cos x) = - R (sin x, cos x)$ : substitute $sin x = t$ .
If $R (- sin x, - cos x) = R (sin x, cos x)$ : substitute $tan x = t$ .

3.2 Linear-in-sin-cos over linear-in-sin-cos

Two very common forms:

$(i) \int \frac{p cos x + q sin x + r}{a cos x + b sin x + c} d x, (ii) \int \frac{p cos x + q sin x}{a cos x + b sin x} d x .$

Rule for (i): Express numerator as $ℓ (denominator) + m (derivative of denominator) + n$ . Find $ℓ, m, n$ by comparing coefficients of $sin x, cos x$ , and the constant. The integral splits into three:

$ℓ \int d x + m \int \frac{( d.c. of denominator )}{denominator} d x + n \int \frac{d x}{a cos x + b sin x + c} .$

Rule for (ii): Express numerator as $ℓ (denominator) + m (derivative of denominator)$ ; find $ℓ, m$ similarly.

Solved Example 4

Evaluate

\int \frac{d x}{sin x ( 2 cos ^{2} x - 1 )} .

Solution:

Check symmetry: replacing $sin x \to - sin x$ changes the sign of the integrand (since $cos x$ is unchanged and $sin x$ appears to the first power). So substitute $cos x = t$ , $d t = - sin x d x$ . Multiplying and dividing by $sin x$ :

$I = \int \frac{sin x d x}{sin ^{2} x ( 2 cos ^{2} x - 1 )} = - \int \frac{d t}{( 1 - t ^{2} ) ( 2 t ^{2} - 1 )} .$

Partial fractions:

$\frac{1}{( 1 - t ^{2} ) ( 2 t ^{2} - 1 )} = \frac{2}{1 - 2 t ^{2}} - \frac{1}{1 - t ^{2}} .$

$I = - \int [\frac{2}{1 - 2 t ^{2}} - \frac{1}{1 - t ^{2}}] d t = - \frac{1}{2} ln \frac{1 + 2 t}{1 - 2 t} + \frac{1}{2} ln \frac{1 - t}{1 + t} + C .$

Back-substitute $t = cos x$ :

$I = - \frac{1}{2} ln \frac{1 + 2 cos x}{1 - 2 cos x} + \frac{1}{2} ln \frac{1 - cos x}{1 + cos x} + C .$

3.3 Integrals of $sin^{M} x cos^{N} x$

For $M, N \in N$ :

If one of $M, N$ is odd, substitute for the term with the even power (i.e. if $M$ is odd, put $cos x = t$ ; if $N$ is odd, put $sin x = t$ ).
If both are odd, either substitution works.
If both are even, use trigonometric identities such as $sin^{2} x = \frac{1 - cos 2 x}{2}$ and $cos^{2} x = \frac{1 + cos 2 x}{2}$ to reduce the powers.

Solved Example 5

Evaluate

\int sin x cos x d x .

Solution:

Put $sin x = t$ , so $cos x d x = d t$ . Then

$I = \int t^{1/2} d t = \frac{2}{3} t^{3/2} + C = \frac{2}{3} (sin x)^{3/2} + C .$

Solved Example 6

Evaluate

\int sin^{2} x cos^{4} x d x .

Solution:

Both powers are even, so use identities:

$sin^{2} x cos^{4} x = (\frac{1 - cos 2 x}{2}) (\frac{1 + cos 2 x}{2})^{2} = \frac{1}{8} (1 - cos 2 x) (1 + cos 2 x)^{2} .$

Expand $(1 + cos 2 x)^{2} = 1 + 2 cos 2 x + cos^{2} 2 x$ :

$sin^{2} x cos^{4} x = \frac{1}{8} (1 - cos 2 x) (1 + 2 cos 2 x + cos^{2} 2 x) .$

Multiplying out and using $cos^{2} 2 x = \frac{1 + cos 4 x}{2}$ and $cos^{3} 2 x = \frac{3 cos 2 x + cos 6 x}{4}$ (the product-to-sum expansion), and integrating term by term, gives after simplification:

$I = \frac{x}{16} + \frac{1}{64} sin 2 x - \frac{1}{64} sin 4 x - \frac{1}{192} sin 6 x + C .$

3.4 Quadratic-in-sin-or-cos Denominators

Integrals of the form

$\int \frac{d x}{a + b sin ^{2} x}, \int \frac{d x}{a + b cos ^{2} x}, \int \frac{d x}{a sin ^{2} x + b cos ^{2} x + c}$

appear in nearly every JEE Mains paper. The universal recipe:

Divide both numerator and denominator by $cos^{2} x$ .
Rewrite $\frac{1}{cos ^{2} x} = sec^{2} x$ and $\frac{sin ^{2} x}{cos ^{2} x} = tan^{2} x$ .
Substitute $tan x = t$ , so $sec^{2} x d x = d t$ .
The result is a rational integral in $t$ of the form $\int \frac{d t}{A t ^{2} + B}$ - a standard $tan^{- 1}$ or $ln$ form.

If the denominator has $sin^{2} x$ but no $cos^{2} x$ , use $sin^{2} x + cos^{2} x = 1$ to insert $cos^{2} x$ before dividing.

Solved Example 7

Evaluate

\int \frac{d x}{3 + 2 sin ^{2} x} .

Solution:

Write $3 = 3 (sin^{2} x + cos^{2} x)$ so the denominator becomes $3 sin^{2} x + 3 cos^{2} x + 2 sin^{2} x = 5 sin^{2} x + 3 cos^{2} x$ . Divide numerator and denominator by $cos^{2} x$ :

$I = \int \frac{sec ^{2} x d x}{5 tan ^{2} x + 3} .$

Put $tan x = t$ , $sec^{2} x d x = d t$ :

$I = \int \frac{d t}{5 t ^{2} + 3} = \frac{1}{5} \int \frac{d t}{t ^{2} + ( 3/5 ) ^{2}} = \frac{1}{5} \cdot \frac{1}{3/5} tan^{- 1} (\frac{t}{3/5}) + C .$

Simplifying $\frac{1}{5 3/5} = \frac{1}{15}$ and back-substituting $t = tan x$ :

$I = \frac{1}{15} tan^{- 1} (\frac{5}{3} tan x) + C .$

Solved Example 8

Evaluate

\int \frac{d x}{4 sin ^{2} x + 5 cos ^{2} x} .

Solution:

Divide numerator and denominator by $cos^{2} x$ :

$I = \int \frac{sec ^{2} x d x}{4 tan ^{2} x + 5} .$

Put $tan x = t$ , $sec^{2} x d x = d t$ :

$I = \int \frac{d t}{4 t ^{2} + 5} = \frac{1}{4} \int \frac{d t}{t ^{2} + ( 5 /2 ) ^{2}} = \frac{1}{4} \cdot \frac{2}{5} tan^{- 1} (\frac{2 t}{5}) + C .$

$I = \frac{1}{2 5} tan^{- 1} (\frac{2 tan x}{5}) + C .$

Common Mistakes to Avoid

Watch out

Skipping the coefficient split $p x + q = ℓ (2 a x + b) + m$ . Trying to integrate $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ by any other method is much harder. The split is the standard approach and expected in JEE solutions.
Forgetting the multiplier when completing the square. For $4 x^{2} + 4 x + 5 = 4 [(x + 1/2)^{2} + 1]$ , the coefficient 4 must be pulled out before applying the standard formula. Omitting it produces an answer off by a factor of $4 = 2$ .
Using Weierstrass when symmetry gives an easier substitution. If the integrand changes sign under $sin x \to - sin x$ , use $cos x = t$ , not $t = tan (x /2)$ . Weierstrass always works but often produces heavy calculations.
Even-power trig integrals: not reducing powers. $\int sin^{4} x d x$ requires $sin^{2} x = (1 - cos 2 x) /2$ applied twice. Trying to substitute $sin x = t$ won't help since the exponent stays intact.
Wrong absolute-value sign in $ln \frac{x - a}{x + a}$ . Some students write it as $ln \frac{a - x}{a + x}$ ; this differs by a factor of $- 1$ inside the log, which flips the sign of the integral. The standard formula requires $x - a$ on top.

Frequently Asked Questions

Q1. What are "integrals of some particular functions" in the JEE syllabus?

This concept covers standard techniques for a small family of frequently-appearing integrand shapes: linear-over-quadratic (with and without square root), rational trigonometric functions $R (sin x, cos x)$ , powers of sine and cosine multiplied together, and irrational algebraic forms like $\frac{1}{( x - k ) ^{r} a x ^{2} + b x + c}$ . Each shape has a standard substitution or algebraic manipulation that reduces it to a table integral.

Q2. How do I integrate $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ ?

Write $p x + q = ℓ (2 a x + b) + m$ , where $2 a x + b$ is the derivative of the quadratic. Find $ℓ$ and $m$ by comparing coefficients. The first part integrates by substitution $u = a x^{2} + b x + c$ (giving $ℓ ln ∣ u ∣$ ); the second part is a standard $\int d x / (a x^{2} + b x + c)$ handled by completing the square.

Q3. What is the Weierstrass substitution and when should I use it?

The Weierstrass or "universal" substitution is $t = tan (x /2)$ , which converts any rational function of $sin x$ and $cos x$ into a rational function of $t$ - integrable by partial fractions. Use it when no obvious symmetry substitution ( $cos x = t$ , $sin x = t$ , or $tan x = t$ ) applies. It's a fallback that always works but often produces heavy algebra.

Q4. How do I decide the substitution for $\int R (sin x, cos x) d x$ ?

Test the symmetry: if replacing $sin x \to - sin x$ flips the sign, use $cos x = t$ ; if $cos x \to - cos x$ flips it, use $sin x = t$ ; if both replaced together leaves it unchanged, use $tan x = t$ ; if none, fall back to $t = tan (x /2)$ .

Q5. What do I do for $\int sin^{M} x cos^{N} x d x$ ?

If one of $M, N$ is odd, substitute for the even-power term: e.g. if $cos x$ has odd power, put $sin x = t$ . If both are odd, either substitution works. If both are even, use the identities $sin^{2} x = (1 - cos 2 x) /2$ and $cos^{2} x = (1 + cos 2 x) /2$ (and product-to-sum) to reduce the powers.

Q6. When do I need to complete the square?

Whenever the denominator is a general quadratic $a x^{2} + b x + c$ (or its square root) with $b \neq = 0$ . Completing the square rewrites it as $a [(x + h)^{2} + k]$ or $a [(x + h)^{2} - k]$ , matching one of the standard $\int d x / (u^{2} \pm a^{2})$ or $\int d x / u^{2} \pm a^{2}$ forms.

Q7. What substitution works for $\int \frac{d x}{( x - k ) ^{r} a x ^{2} + b x + c}$ ?

Put $x - k = \frac{1}{t}$ , so $d x = - \frac{d t}{t ^{2}}$ . This transforms the integral into $\int \frac{t ^{r - 1} d t}{A t ^{2} + B t + C}$ for some constants $A, B, C$ - a form treatable by completing the square.

Q8. Are all these standard forms testable in JEE Mains and Advanced?

Yes. Every category listed here appears regularly in both JEE Mains and JEE Advanced. In Mains, the integrals are usually one-step applications of the standard technique; in Advanced, they may be nested inside larger problems (definite integrals, area calculations, or differential equations).

Previous year questions on Integrals of Some Particular Functions

1 question from past papers, each with a step-by-step solution.

JEE Main 2025 Apr 2 Shift 2, Mathematics Q13

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1. Algebraic Integrals with Linear Numerator

Ratio of two quadratics

2. Integration of Irrational Algebraic Fractions

3. Trigonometric Integrals

3.1 Integrals of R(sinx,cosx)

3.2 Linear-in-sin-cos over linear-in-sin-cos

3.3 Integrals of sinMxcosNx

3.4 Quadratic-in-sin-or-cos Denominators

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What are "integrals of some particular functions" in the JEE syllabus?

Q2. How do I integrate ∫ax2+bx+cpx+q​dx?

Q3. What is the Weierstrass substitution and when should I use it?

Q4. How do I decide the substitution for ∫R(sinx,cosx)dx?

Q5. What do I do for ∫sinMxcosNxdx?

Q6. When do I need to complete the square?

Q7. What substitution works for ∫(x−k)rax2+bx+c​dx​?

Q8. Are all these standard forms testable in JEE Mains and Advanced?

Previous year questions on Integrals of Some Particular Functions

Ready to master Integrals?

3.1 Integrals of $R (sin x, cos x)$

3.3 Integrals of $sin^{M} x cos^{N} x$

Q2. How do I integrate $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ ?

Q4. How do I decide the substitution for $\int R (sin x, cos x) d x$ ?

Q5. What do I do for $\int sin^{M} x cos^{N} x d x$ ?

Q7. What substitution works for $\int \frac{d x}{( x - k ) ^{r} a x ^{2} + b x + c}$ ?