# Inter 2nd Year Maths 2B Integration Formulas

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 6 Integration to solve questions creatively.

## Intermediate 2nd Year Maths 2B Integration Formulas

→ Integration is the inverse process of differentiation.

→ Let A ⊆ R and let f: A → R be a function. If there is a function B on A such that F'(x) = f(x), ∀ x ∈ A, then we call B an antiderivative of for a primitive of f.
i.e., $$\frac{d}{d x}$$(sin x) = cos x, ∀ x ∈ R ax
f(x) = cos x, x ∈ R, then the function
F(x) = sin x, x ∈ R is an antiderivative or primitive of f.

→ If F is an antiderivative offon A, then for k ∈ R, we have (F + k) (x) = f(x), ∀ x ∈ A.

→ Hence F + k is also an antiderivative off.
∴ c is any real number F(x + c) = G(x) = sin x + c, ∀ x ∈ R is also an antiderivative of cos x.

→ It is denoted by ∫ (cos x) dx = sin x + c, (i.e.) ∫ f(x) dx = F(x) + c.

→ Here c is called a constant of integration,
f is called the integrand and x is called the variable of integration.

Standard Forms:

→ ∫xn dx = $$\frac{x^{n+1}}{n+1}$$ + c if n ≠ – 1

→ ∫$$\frac{1}{x}$$ dx = log |x| + c

→ ∫ sin x dx = – cos x + c, x ∈ R

→ ∫ cos x dx – sin x + c, x ∈ R

→ ∫tan x dx = log |sec x| + c

→ ∫ cot x dx = log |sin x | + c

→ ∫sec x dx = log |sec x + tan x | + c (or) log |tan $$\left(\frac{\pi}{4}+\frac{x}{2}\right)$$| + c

→ ∫cosec x dx = log |cosec x – cot x| + c (or) log |tan$$\left(\frac{x}{2}\right)$$| + c

→ ∫sec2 x dx = tan x + c, x ∈ R – $$\left|\frac{n \pi}{2}\right|$$, n is odd integer

→ ∫cosec2 x dx = – cot x + c → R – nπ, n ∈ Z

→ ∫sec x tan x dx = sec x + c, R – $$\left[\frac{n \pi}{2}\right]$$, n is an odd integer

→ ∫cosec x cot xdx = – cosec x + c, R – [nπ], n ∈ Z

→ ∫ex dx = ex + c, x ∈ R

→ ∫ax dx = $$\frac{a^{x}}{\log _{e} a}$$ + c, a > 0, a ≠ 1

→ ∫$$\frac{1}{\sqrt{1+x^{2}}}$$ dx = sin-1x + C = – cos-1 (x) + c

→ ∫$$\frac{d x}{1+x^{2}}$$ dx = tan-1x + C = – cot-1 (x) + c

→ ∫$$\frac{d x}{|x| \sqrt{x^{2}-1}}$$ dx = sec-1x + C = – cosec-1 (x) + c

→ ∫ sinh x dx = cosh x + c

→ ∫cosh xdx = sinh x + c

→ ∫cosec2h x dx coth x + c

→ ∫sec2h x dx = tanh x + c

→ ∫cosech x coth xdx = – cosech x + c

→ ∫sech x tanh x dx = – sech x + c

→ ∫eax dx = $$\frac{e^{a x}}{a}$$ + c

→ ∫eax+b dx = $$\frac{e^{a x+b}}{a}$$ + c

→ ∫sin (ax + b) dx = \frac{-\cos (a x+b)}{a} + c

→ ∫cos (ax + b) dx = $$\frac{\sin (a x+b)}{a}$$ + c

→ ∫sec2 (ax + b) dx = $$\frac{\tan (a x+b)}{a}$$ + c

→ ∫cosec2 (ax + b) dx = $$\frac{-\cot (a x+b)}{a}$$ + c

→ ∫cosec(ax + b) cot(ax + b) dx = $$\frac{-{cosec}(a x+b)}{a}$$ + c

→ ∫sec (ax + b) tan(ax + b) dx = $$\frac{\sec (a x+b)}{a}$$ + c

→ ∫f(x).g(x) dx = f(x) ∫g(x) dx – ∫[$$\frac{d}{d x}$$ f(x) . ∫ g(x) dx] dx (called as integration by parts)

→ ∫$$\frac{1}{\sqrt{1+x^{2}}}$$ dx = sinh-1 (x) + c, x ∈ R = log (x + $$\sqrt{x^{2}+1}$$) + c, x ∈ R

→ ∫$$\frac{1}{\sqrt{x^{2}-1}}$$ dx = cosh-1 (x) + c (or) log (x + $$\sqrt{x^{2}-1}$$) + c, x ∈ (1, ∞)
= – cos h-1 (- x) + c (or) log (x + $$\sqrt{x^{2}-1}$$) + c, x ∈ (1, ∞)
= log |x + $$\sqrt{x^{2}-1}$$| + c, x ∈ R – [- 1, 1]

→ ∫ex [f(x) + f'(x)] dx = ex. f(x) + c

→ ∫$$\frac{1}{\sqrt{a^{2}-x^{2}}}$$ dx = sin-1$$\left(\frac{x}{a}\right)$$ + c

→ ∫$$\frac{1}{\sqrt{a^{2}+x^{2}}}$$ dx = sinh-1$$\left(\frac{x}{a}\right)$$ + c (or) log $$\frac{\left|x+\sqrt{x^{2}+a^{2}}\right|}{a}$$ + c

→ ∫$$\frac{1}{\sqrt{x^{2}-a^{2}}}$$ dx = cosh-1$$\left(\frac{x}{a}\right)$$ + c (or) log $$\frac{\left|x+\sqrt{x^{2}-a^{2}}\right|}{a}$$ + c

→ ∫$$\frac{1}{a^{2}+x^{2}}$$ dx = $$\frac{1}{a}$$ tan-1$$\left(\frac{x}{a}\right)$$+ c

→ ∫$$\frac{1}{\sqrt{a^{2}-x^{2}}}$$ = $$\frac{1}{2a}$$ log $$\left|\frac{a+x}{a-x}\right|$$ + C

→ ∫$$\frac{1}{x^{2}-a^{2}}$$ dx = $$\frac{1}{2a}$$ log $$\left|\frac{x-a}{x+a}\right|$$ + c

→ ∫$$\sqrt{a^{2}-x^{2}}$$ dx = $$\frac{1}{2}$$x $$\sqrt{a^{2}-x^{2}}$$ + $$\frac{a^{2}}{2}$$ sin-1$$\left(\frac{x}{a}\right)$$ + c

→ ∫$$\sqrt{x^{2}+a^{2}}$$ dx = $$\frac{1}{2}$$x $$\sqrt{x^{2}+a^{2}}$$ + $$\frac{a^{2}}{2}$$ sinh-1$$\left(\frac{x}{a}\right)$$ + c

→ ∫$$\sqrt{x^{2}-a^{2}}$$ dx = $$\frac{1}{2}$$x $$\sqrt{x^{2}-a^{2}}$$ + $$\frac{a^{2}}{2}$$ cosh-1$$\left(\frac{x}{a}\right)$$ + c

→ To evaluate

• $$\frac{p x+q}{a x^{2}+b x+c}$$ dx
• ∫ (px + q) $$\sqrt{a x^{2}+b x+c}$$ dx
• ∫$$\frac{p x+q}{\sqrt{a x^{2}+b x+c}}$$ dx, where a, b, c, p, q ∈ R write
px + q = A.$$\frac{d}{d x}$$ (ax2 + bx + c) + B and then integrate.

→ To evaluate ∫$$\frac{d x}{(a x+b) \sqrt{p x+q}}$$ where a, b, c, p, q, ∈ R put t2 = px + q

→ To evaluate ∫$$\frac{1}{a+b \cos x}$$ dx (or) $$\frac{1}{a+b \sin x}$$ dx
(or) $$\frac{1}{a+b \cos x+c \sin x}$$ dx, put tan $$\frac{x}{2}$$ = t
Then sin x = $$\frac{2 t}{1+t^{2}}$$, cos x = $$\frac{1-t^{2}}{1+t^{2}}$$ and dx = $$\frac{2}{1+t^{2}}$$ dt

→ To evaluate ∫$$\frac{a \cos x+b \sin x+c}{d \cos x+e \sin x+f}$$ dx where a, b, c, d e, f ∈ R; d ≠ 0, e ≠ 0, write a cos x + b sin x + c = A [d cos x + e sin x + f]’ + B (d cos x + e sin x + f) + ∨.
Find A, B, ∨ and then integrate.

→ If In = ∫xn . eax dx then In = $$\frac{x^{n} \cdot e^{a x}}{a}-\frac{n}{a}$$ In – 1 for n ∈ N

→ If In = ∫ sinn (x) dx then In = – $$\frac{\sin ^{n-1}(x) \cos x}{n}$$ + $$\left(\frac{n-1}{n}\right)$$ In – 2 for n ∈ N, n ≥ 2

→ f In = ∫ cosn (x) dx then In = – $$\frac{\cos ^{n-1}(x) \sin x}{n}$$ + $$\left(\frac{n-1}{n}\right)$$ In – 2 for n ∈ N, n ≥ 2

→ If In = ∫tann (x) dx then In = $$\frac{\tan ^{n-1}(x)}{n-1}$$ In – 2 for N ∈ n, n ≥ 2

→ If Im, n = ∫ sinm (x) cosn (x) dx then
If Im, n = $$\frac{1}{m+n}$$ cosn – 1 (x) sinm + 1 (x) + $$\left(\frac{n-1}{m+n}\right)$$ Im, n – 2 where m, n ∈ N, n ≥ 2

→ If Im, n = ∫secn (x) dx then In = $$\frac{\sec ^{n-2}(x) \tan x}{n-1}$$ + $$\left(\frac{n-2}{n-1}\right)$$ In – 2

Theorem: If f(x) and g(x) are two integrable functions then
∫ f(x).g(x)dx = f(x)∫g(x)dx – ∫f’(x)[∫g(x)dx] dx.
Proof:
$$\frac{\mathrm{d}}{\mathrm{dx}}$$ [f(x). ∫g(x)dx] = f(x) $$\frac{\mathrm{d}}{\mathrm{dx}}$$[∫ g(x)dx] + ∫g(x)dx .$$\frac{\mathrm{d}}{\mathrm{dx}}$$[f(x)]
= f(x)g(x) + [∫g(x)dx]f’(x)
∴ ∫[f(x)g(x) + f’(x)∫g(x)dx] dx = f(x)∫g(x)dx
⇒ ∫f (x)g(x)dx + ∫f’(x) [∫g(x)dx] dx = f (x)∫g(x) dx
∴ ∫f(x)g(x)dx = f(x)∫g(x)dx – ∫f’(x)[∫g(x)dx]dx

Note 1: If u and v are two functions of x then ∫u dv = uv – ∫v du.

Note 2: If u and v are two functions of x; u’, u”, u”’ …………. denote the successive derivatives of u and v1, v2, v3, v4, v5 … the successive integrals of v then the extension of integration by pairs is
∫uv dx = uv1 – u’v2 + u”v3 – u”’v4 + ………

Note 3: In integration by parts, the first function will be taken as the following order.
Inverse functions, Logarithmic functions, Algebraic functions, Trigonometric functions and Exponential functions. (To remember this a phrase ILATE).

Theorem: ∫eax cos bx dx = $$\frac{e^{a x}}{a^{2}+b^{2}}$$ (a cos bx + b sin bx) + c
Proof:

Theorem: ∫eax sin bx dx = $$\frac{e^{a x}}{a^{2}+b^{2}}$$ (a sin bx – b cos bx)
Proof:
Let I = ∫eax sin bx dx = sin bx ∫eax dx – ∫[d(sin bx) ∫eax dx] dx

Theorem: ∫ ex [f(x) + f’(x)]dx = exf(x) + c
Proof:
∫ex [f(x) + f’(x)]dx = ∫ex f(x)dx + ∫ex f’(x)dx
= f(x) ∫ exdx – ∫[d[f(x)] ∫exdx] dx + ∫ex f'(x)dx
= f(x)ex – ∫f'(x)exdx + ∫exf'(x) dx = exf(x) + c
Note: ∫e-x [f(x) – f’(x)]dx = – e-xf(x) + c

Definition: If f(x) and g(x) are two functions such that f’(x) = g(x) then f(x) is called antiderivative or primitive of g(x) with respect to x.

Note 1: If f(x) is an antiderivative of g(x) then f(x) + c is also an antiderivative of g(x) for all c ∈ R.

Definition: If F(x) is an antiderivative of f(x) then F(x) + c, c ∈ R is called indeVinite integral of f(x) with respect to x. It is denoted by ∫f(x)dx. The real number c s called constant of integration.

Note:

• The integral of a function need not exists. If a function f(x) integral then f(x) is called an integrable function.
• The process of finding the integral of a function is known as Integration.
• The integration is the reverse process of differentiation.

Corollary:
If f(x), g(x) are two integrable functions then ∫(f ± g) (x) dx = ∫f(x)dx ± ∫fg(x)dx

Corollary:
If f1(x), f2(x), ……, fn(x) are integrable functions then
∫(f1 + f2 + …….. + fn)(x)dx = ∫f1(x)dx + ∫f2(x)dx + ……. + ∫fn(x)dx.

Corollary:
If f(x), g(x) are two integrable functions and k, l are two real numbers then ∫(kf + lg) (x)dx = k∫f(x) dx + 1∫g(x)dx.

Theorem: If f f(x)dx = g(x) and a ≠ 0 then ∫ f(ax + b)dx = $$\frac{1}{a}$$g(ax+b)+c.
Proof:
Put ax + b = t.

Theorem: It f(x) is a differentiable function then ∫$$\frac{f^{\prime}(x)}{f(x)}$$ dx = log |f(x)| + c.
Proof:
Put f(x) = t ⇒ f’(x) = $$\frac{d \mathrm{t}}{\mathrm{dx}}$$ ⇒ f’(x)dx = dt
∴ ∫$$\frac{f^{\prime}(x)}{f(x)}$$ = ∫latex]\frac{1}{\mathrm{t}}[/latex] dt = log |t| + c = log |f(x)| + c

Theorem: ∫tan x dx = log |sec x| for x ≠ (2n + 1)$$\frac{\pi}{2}$$, n ∈ Z.
Proof:
∫tan x dx = ∫$$\frac{\sin x}{\cos x}$$ dx = -∫$$\frac{d(\cos x)}{\cos x}$$ dx
= – log |cos x| + c = log$$\frac{1}{|\cos x|}$$ + c = log|sec x| + c

Theorem: ∫cot x dx = log |sin x| + c for x ≠ nπ, n ∈ Z.
Proof:
∫cot x dx = ∫$$\frac{\cos x}{\sin x}$$ dx = log |sin x| + c

Theorem: ∫ sec x dx = log |sec x + tan x| + c = log |tan(π/4 + x/2) + c for x ≠ (2n + 1)$$\frac{\pi}{2}$$, n ∈ Z.
Proof:

Theorem: ∫csc x dx = log|csc x – cot x| + c = log |tan x/2| + c for x ≠ nπ, n ∈ Z.
Proof:
∫csc x dx = $$\int \frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}$$ dx
= $$\int \frac{\csc ^{2} x-\csc x \cot x}{\csc x-\cot x}$$ dx = log |csc x – cot x| + c
= log$$\left|\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right|$$ + c
= log$$\left|\frac{1-\cos x}{\sin x}\right|$$ + c
= log$$\left|\frac{2 \sin ^{2} x / 2}{2 \sin x / 2 \cos x / 2}\right|$$ + c
= log |tan x/2| + c

Theorem: If f(x) is differentiable function and n ≠ – 1 then ∫[f(x)]n f’(x)dx = $$\frac{[f(x)]^{n+1}}{n+1}$$ + c.
Proof:
Put f(x) = t ⇒ f’(x) dx = dt

Theorem: If ∫f(x)dx = F(x) and g(x) is a differentiable function then ∫ (fog)(x)g’(x) dx = F[g(x)] + c.
Proof:
g(x) = t ⇒ g’(x) dx = dt
∴ ∫(fog)(x)g’(x)dx = ∫f[g(x)]g’(x) dx
= ∫f(t)dt = F(t) + c = F[g(x)] + c

Theorem: ∫$$\frac{1}{\sqrt{a^{2}-x^{2}}}$$ dx = Sin-1 + c for x ∈ (- a, a)
Proof:
Put x = a sin θ. Then dx = a cos θ dθ

Theorem: ∫$$\frac{1}{\sqrt{a^{2}+x^{2}}}$$dx = Sinh-1  + c for x ∈ R.
Proof:
Put x = a sinhθ. Then dx = a cos hθ dθ
∴ ∫$$\frac{1}{\sqrt{a^{2}+x^{2}}}$$ dx = $$\int \frac{1}{\sqrt{a^{2}+a^{2} \sinh ^{2} \theta}}$$ a coshθ dθ
= ∫$$\frac{a \cosh \theta}{a \cosh \theta}$$ = ∫dθ = θ + c = Sinh-1$$\left(\frac{x}{a}\right)$$ + c

Theorem:
∫dx = Cosh-1 + c for x ∈ (- ∞, – a) ∪ (a, ∞)
Proof:
Put x = a coshθ. Then dx = a sin hθ dθ
∴ ∫$$\frac{1}{\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}}$$ dx = ∫$$\frac{1}{\sqrt{a^{2} \cosh ^{2} \theta-a^{2}}}$$ a sin hθ dθ
= ∫$$\frac{a \sinh \theta}{a \sinh \theta}$$ dθ = ∫ dθ = θ + c = Cosh-1$$\left(\frac{x}{a}\right)$$ + c

Theorem:
∫$$\frac{1}{a^{2}+x^{2}}$$ dx = $$\frac{1}{a}$$ Tan-1$$\left(\frac{x}{a}\right)$$ + c for x ∈ R.
Proof:
Put x = a tan θ. Then dx = a sec2θ dθ

Theorem:
∫$$\frac{1}{a^{2}-x^{2}}$$dx = $$\frac{1}{2 \mathrm{a}}$$ log$$\left|\frac{a+x}{a-x}\right|$$ + c for x ≠ ± a
Proof:
∫$$\frac{1}{a^{2}-x^{2}}$$dx = ∫$$\left|\frac{a+x}{a-x}\right|$$dx
= $$\frac{1}{2 \mathrm{a}} \int\left(\frac{1}{\mathrm{a}+\mathrm{x}}+\frac{1}{\mathrm{a}-\mathrm{x}}\right)$$dx = $$\frac{1}{2 \mathrm{a}}$$ [log |a + x| – log |a – x|] + c
= $$\frac{1}{2 \mathrm{a}}$$ log$$\left|\frac{a+x}{a-x}\right|$$ + c

Theorem:
∫$$\frac{1}{x^{2}-a^{2}}$$ dx = $$\frac{1}{2 \mathrm{a}}$$ log $$\left|\frac{x-a}{x+a}\right|$$ + c for x ≠± a
Proof:
∫$$\frac{1}{x^{2}-a^{2}}$$ dx = ∫$$\frac{1}{(x-a)(x+a)}$$ dx
= $$\frac{1}{2 a} \int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$$ dx = $$\frac{1}{2 \mathrm{a}}$$ [log |x – a| – log |x + a|] + c
= $$\frac{1}{2 \mathrm{a}}$$ log $$\left|\frac{x-a}{x+a}\right|$$ + c

Theorem:
∫$$\sqrt{a^{2}-x^{2}}$$dx = $$\frac{x}{2} \sqrt{a^{2}-x^{2}}$$ + $$\frac{\mathrm{a}^{2}}{2}$$ sin-1$$\left(\frac{x}{a}\right)$$ + c for x ∈ (- a, a)
Proof:
Put x = a sin θ. Then dx = a cos θ dθ

Theorem:
∫$$\sqrt{a^{2}+x^{2}}$$ dx = $$\frac{x}{2} \sqrt{a^{2}+x^{2}}$$ + $$\frac{\mathrm{a}^{2}}{2}$$ Sinh-1 $$\left(\frac{\mathrm{x}}{\mathrm{a}}\right)$$ + c for x ∈ R.
Proof:
Put x = sinhθ. Then dx = a coshθ dθ
∴ ∫$$\sqrt{a^{2}+x^{2}}$$ dx = ∫$$\sqrt{a^{2}+a^{2} \sinh ^{2} \theta}$$ a coshθ dθ
= ∫$$a \sqrt{1+\sinh ^{2} \theta}$$ a coshθ dθ = a2 ∫cosh2 θdθ
= $$=\mathrm{a}^{2} \int \frac{1+\cosh 2 \theta}{2} \mathrm{~d} \theta=\frac{\mathrm{a}^{2}}{2}\left[\theta+\frac{1}{2} \sinh 2 \theta\right]+\mathrm{c}$$
= $$\frac{a^{2}}{2}\left[\theta+\frac{1}{2} 2 \sinh \theta \cosh \theta\right]+c$$
= $$\frac{a^{2}}{2}\left[\theta+\sinh \theta \sqrt{1+\sinh ^{2} \theta}\right]+c$$
= $$\frac{a^{2}}{2}\left[{Sinh}^{-1}\left(\frac{x}{a}\right)+\frac{x}{a} \sqrt{1+\frac{x^{2}}{a^{2}}}\right]+c$$
= $$\frac{\mathrm{a}^{2}}{2}{Sinh}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\frac{\mathrm{x}}{\mathrm{a}} \sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}+\mathrm{c}$$

Theorem:
∫$$\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}$$ dx = $$\frac{x}{2} \sqrt{x^{2}-a^{2}}$$ – $$\frac{a^{2}}{2}$$ Cosh-1$$\left(\frac{x}{a}\right)$$ + c for x ∈ [a, ∞)
Proof:
Put x = a coshθ. Then dx = a sinhθ dθ