Methods of integration: substitution and identities

What is integration by substitution?

In plain words: substitution is the chain rule of NCERT Chapter 5 read backwards. (The definition follows NCERT’s x = g(t); in practice we name the new variable directly, t = g(x), as NCERT’s examples do.) If part of the integrand is a function t = g(x) and its derivative g′(x) is also there as a factor, then g′(x) dx becomes dt, and the integral turns into a standard one in t. NCERT’s advice (page 236): usually substitute for a function whose derivative also occurs in the integrand. At the end, put x back.

NCERT Section 7.3 names three methods: substitution, partial fractions and integration by parts (page 235). This page covers substitution and the trigonometric identities of Section 7.3.2.

Choosing the substitution

NCERT Example 5 (pages 236–237) shows the pattern:

• For m ≠ 0, ∫sin mx dx: put t = mx, so dx = dt/m. The integral is (1/m)∫sin t dt = −(1/m) cos mx + C. If m = 0, the integrand is 0 and the answer is C.
• ∫2x sin (x² + 1) dx: put t = x² + 1, so 2x dx = dt. The integral is −cos (x² + 1) + C.
• ∫sin (tan^⁻¹ x)/(1 + x²) dx: put t = tan^⁻¹ x, so dx/(1 + x²) = dt. The integral is −cos (tan^⁻¹ x) + C.

Sometimes two substitutions are needed. For ∫(tan⁴ √x sec² √x)/√x dx, NCERT puts t = √x, so dx = 2t dt and the integral is 2∫tan⁴ t sec² t dt; then u = tan t gives 2∫u⁴ du:

∫(tan⁴ √x sec² √x)/√x dx

= (2/5) tan⁵ √x + C.

Two patterns come straight from the rule (reasoned from NCERT’s substitution; NCERT uses the first for tan x and cot x on pages 237–238): with t = f(x),

Integrals of tan x, cot x, sec x and cosec x

NCERT derives these by substitution on pages 237–238 and uses them later without reference:

• tan x: write it as sin x/cos x and put t = cos x, so sin x dx = −dt. The integral is −log |cos x| + C, which is log |sec x| + C.
• cot x: cos x/sin x with t = sin x gives log |sin x| + C.
• sec x: multiply the top and bottom by sec x + tan x. The new top, sec x(sec x + tan x), is the derivative of the bottom, so t = sec x + tan x gives log |sec x + tan x| + C.
• cosec x: multiply by cosec x + cot x instead. Now dt is −cosec x times (cosec x + cot x) dx, so the integral is −log |cosec x + cot x| + C. Since cosec² x − cot² x = 1, the product of cosec x + cot x and cosec x − cot x is 1, so this is log |cosec x − cot x| + C.

Beyond the NCERT text (JEE): the integral of sec x can also be written as log |tan (π/4 + x/2)| + C, and that of cosec x as log |tan (x/2)| + C. These are the same answers, since tan (π/4 + x/2) equals sec x + tan x, and tan (x/2) equals cosec x − cot x.

Using trigonometric identities

When the integrand is a power or a product of sines and cosines, an identity first turns it into a sum of terms that can be integrated one by one (NCERT Section 7.3.2, pages 241–242). The identities are the Class 11 double angle formulas and the product-to-sum formulas. NCERT Example 7:

• cos² x = (1 + cos 2x)/2, so ∫cos² x dx = x/2 + (1/4) sin 2x + C.
• sin 2x cos 3x = ½[sin 5x + sin (−x)] = ½(sin 5x − sin x), so the integral is −(1/10) cos 5x + ½ cos x + C.
• sin³ x = (3 sin x − sin 3x)/4, so ∫sin³ x dx = −(3/4) cos x + (1/12) cos 3x + C.

NCERT also finds ∫sin³ x dx by substitution: write sin³ x = (1 − cos² x) sin x and put t = cos x, which gives −cos x + (1/3) cos³ x + C. The two answers look different, but NCERT’s Remark (page 242) notes that they are equivalent: putting cos 3x = 4 cos³ x − 3 cos x into the first one gives exactly the second.

Solved examples

Common mistakes to avoid

Quick revision summary

• Substitution: put t = g(x) when g′(x) is also a factor of the integrand, so that g′(x) dx = dt; change dx too, and put x back at the end.
• ∫f′(x)/f(x) dx = log |f(x)| + C; this gives ∫tan x dx = log |sec x| + C and ∫cot x dx = log |sin x| + C.
• ∫sec x dx = log |sec x + tan x| + C and ∫cosec x dx = log |cosec x − cot x| + C.
• Powers and products of sines and cosines: use the double angle, triple angle and product-to-sum identities first, such as cos^² x = (1 + cos 2x)/2.

Questions students ask

Why does integration by substitution work?

Integration by substitution works because it is the chain rule read backwards: if F′ = f, then F(g(x)) has derivative f(g(x)) g′(x), so that product integrates to F(g(x)) + C; putting t = g(x) is shorthand for this. In ∫x² cos (x³) dx, t = x³ gives dt = 3x² dx and the answer (1/3) sin (x³) + C, which the chain rule turns back into the integrand. A missing constant can be supplied, like the 1/3, but not a missing variable: in ∫cos (x³) dx, the same t leaves dx = dt/(3x²), still with x.

What is the difference between integration by substitution and integration by parts?

Integration by substitution undoes the chain rule and suits an integrand containing a function and its derivative as a factor, while integration by parts undoes the product rule and suits a product of two kinds of function, one of which gets simpler when differentiated, such as x times e^x. So look first for an inner function and its derivative: if both are there, substitute. Some integrals need both: in ∫e^√x dx (x > 0), t = √x gives dx = 2t dt and 2∫te^t dt, which parts turns into 2e^t(t − 1), so the answer is 2e^√x(√x − 1) + C.

How do you integrate powers of sin x and cos x?

To integrate sin^m x cosⁿ x, for nonnegative integer powers m, n, look at the powers: if one is odd, set one factor of that function aside, turn the rest into the other function with sin² x + cos² x = 1 and put t equal to the other function; if both are even, lower the powers with double angle formulas such as cos² x = (1 + cos 2x)/2. For ∫sin⁵ x dx, write sin⁴ x as (1 − cos² x)² and put t = cos x: the integral is −cos x + (2/3) cos³ x, minus (1/5) cos⁵ x, plus C. NCERT’s Example 6 (page 238) does sin³ x cos² x this way.

What is the integral of sec x cosec x?

The integral of sec x cosec x, that is of 1/(sin x cos x), is log |tan x| + C, on any interval where sin x and cos x are both non-zero. Dividing the top and the bottom by cos² x turns the integrand into sec² x/tan x, whose top is the derivative of its bottom, so t = tan x gives ∫dt/t = log |t| + C. Another route writes 1 = sin² x + cos² x in the top, which leaves tan x + cot x, and the standard integrals log |sec x| and log |sin x| add up to the same log |tan x|.

What is trigonometric substitution in integration?

Trigonometric substitution puts x = a sin θ, a tan θ or a sec θ (a > 0), so that √(a² − x²), a² + x² or √(x² − a²) becomes a cos θ, a² sec² θ or a tan θ (choose −π/2 ≤ θ ≤ π/2 for sine, −π/2 < θ < π/2 for tangent, and 0 ≤ θ < π/2 for secant on x ≥ a; exclude endpoints that make an integrand undefined); NCERT uses it to prove the particular integrals and, on page 265, names it as another route to the square-root integrals. For ∫dx/(x² + 9)^3/2, x = 3 tan θ leaves (1/9)∫cos θ dθ = (1/9) sin θ + C, and sin θ = x/√(x² + 9), giving x/(9√(x² + 9)) + C.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find ∫(log x)²/x dx and ∫dx/(x + x log x) (NCERT Exercise 7.2, Q2 and Q3, page 240).

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   With t = log x and dt = dx/x: ∫t² dt = (1/3)(log x)³ + C. For the second, x + x log x = x(1 + log x); put t = 1 + log x to get log |1 + log x| + C.

2. Q2. Find ∫e^{2x + 3} dx and ∫x/e^x2 dx (NCERT Exercise 7.2, Q16 and Q17, page 240).

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   ½e^{2x + 3} + C. For the second, x/e^x2 = xe^−x2; put t = −x², so x dx = −½ dt, giving −½e^−x2 + C.

3. Q3. Find ∫sin² (2x + 5) dx (NCERT Exercise 7.3, Q1, page 243).

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   sin² θ = (1 − cos 2θ)/2 with θ = 2x + 5, so the integrand is ½ − ½ cos (4x + 10), and the integral is x/2 − (1/8) sin (4x + 10) + C.

4. Q4. With the top 10x⁹ + 10^x log 10 and the bottom x¹⁰ + 10^x, the integral of top/bottom equals (A) 10^x − x¹⁰ + C; (B) 10^x + x¹⁰ + C; (C) (10^x − x¹⁰)⁻¹ + C; (D) log (10^x + x¹⁰) + C (NCERT Exercise 7.2, Q38, page 241).

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   The top is the derivative of the bottom, so the integral is log |x¹⁰ + 10^x| + C. The bottom is positive, so this is (D).