TS Inter 2nd Year Maths 2B Integration Formulas

Learning these TS Inter 2nd Year Maths 2B Formulas Chapter 6 Integration will help students to solve mathematical problems quickly.

TS Inter 2nd Year Maths 2B Integration Formulas

→ A function F(x) Is called an anti derivative or indefinite Integral of a function f(x) If \(\frac{d}{d x}\)[F (x)] = f(x).

→ Indefinite Integrals:
Let f(x) be a function. Then the collection of all its anti derivatives is called the indefinite integral of f(x) and it is denoted by ∫f(x) dx.
Thus \(\frac{d}{d x}\) [F(x) + c] = f(x) ⇔ ∫f(x)dx = F(x) + c
where F(x) is the anti derivative and c is the arbitrary constant known as the constant of integration.

→ Standard forms:
∫K dx = Kx + c, K is constant

→ ∫1 dx = x + c

→ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + c, n ≠ -1

→ ∫x dx = \(\frac{x^2}{2}\) + c

→ ∫√x dx = \(\frac{2}{3}\)x√x + c

→ ∫\(\frac{1}{\sqrt{x}}\)dx = 2√x + c

→ ∫\(\frac{1}{x^2}\)dx = \(\frac{-1}{x}\) + c

→ ∫\(\frac{1}{x}\)dx = log |x| + c

→ ∫e^xdx = e^x + c

→ ∫a^xdx = \(\frac{a^x}{\log a}\) + c

→ ∫sin x dx = – cos x + c

→ ∫cos x dx = sin x + c

→ ∫tan x dx = log|sec x| + c = -log |cos x| + c

→ ∫cot x dx = log|sin x| + c = -log|cosec x| + c

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

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

→ ∫sec²x dx = tan x + c

→ ∫cosec²x dx = -cot x + c

→ ∫sec x tan x dx = sec x + c

→ ∫cosec x cot x dx = -cosec x + c

→ ∫sin hx dx = cos hx + c

→ ∫cos hx dx = sin hx + c

→ ∫tan hx dx = log|cos hx| + c

→ ∫cot hx dx = log |sin hx| + c

→ ∫sech²x dx = tan hx + c

→ ∫cosech²x dx = -cot hx + c

→ ∫sec hx tan hx dx = -sec hx + c

→ ∫cosec hx cot hx dx = – cosec hx + c

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

→ ∫\(\frac{1}{\sqrt{x^2-1}}\) dx = cosh^-1x + c = log(x + \(\sqrt{x^2-1}\)) + c

→ ∫\(\frac{1}{\sqrt{x^2+1}}\) dx = sin h^-1x + c = log(x + \(\sqrt{x^2+1}\)) + c

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

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

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

→ ∫\(\frac{1}{x \sqrt{x^2-1}}\)dx = sec^-1x + c

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

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

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

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

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

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

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

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

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

→ \(\frac{d}{d x}\)[∫f(x)dx] = f(x)

→ ∫kf(x) dx = k.∫f(x)dx

→ ∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx

→ ∫\(\frac{f^{\prime}(x)}{f(x)}\)dx = log|f(x)| + c

→ ∫\(\frac{f^{\prime}(x)}{\sqrt{f(x)}}\)dx = 2\(\sqrt{f(x)}\) + c

→ ∫[f(x)]ⁿ.f'(x)dx = \(\frac{f^{n+1}(x)}{n+1}\) + c, n ≠ -1

→ ∫\(\frac{f^{\prime}(x)}{1+[f(x)]^2}\)dx = tan^-1f(x) + c

→ Evaluation of Integral of various types by using standard results:

Type of intrgral	Technique (or) Substitution
1. ∫\(\frac{1}{a x^2+b x+c}\)dx (or) ∫\(\frac{1}{\sqrt{a x^2+b x+c}}\)dx (or) ∫\(\sqrt{a x^2+b x+c}\)dx	Express ax2 + bx + c as the sum or difference of two squares. May he reduced to one of the forms like ∫\(\frac{1}{a^2+x^2}\) dx (or) ∫\(\frac{1}{a^2-x^2}\) dx (or) ∫\(\frac{1}{x^2-a^2}\) dx (or) ∫\(\frac{1}{\sqrt{a^2-x^2}}\) dx (or) ∫\(\frac{1}{\sqrt{x^2-a^2}}\) dx (or) ∫\(\frac{1}{\sqrt{a^2+x^2}}\) dx (or) ∫\(\sqrt{a^2+x^2}\) dx (or) ∫\(\sqrt{a^2-x^2}\) dx (or) ∫\(\sqrt{x^2-a^2}\) dx
2. ∫\(\frac{p x+q}{a x^2+b x+c}\)dx (or) ∫\(\frac{p x+q}{\sqrt{a x^2+b x+c}}\)dx (or) ∫(px + q)\(\sqrt{a x^2+b x+c}\)dx	Write px + q = A\(\frac{d}{d x}\)(ax2 + bx + c) + B where A, B are constants to be determined by equating the coefficients of similar terms on both sides.
3. ∫\(\frac{1}{(p x+q) \sqrt{a x^2+b x+c}}\) dx	Put px + q = \(\frac{1}{t}\)
4. ∫\(\frac{1}{\left(A x^2+B\right)\left(C x^2+D\right)}\)dx	Put x = \(\frac{1}{t}\)
5. ∫\(\sqrt{(x-\alpha)(\beta-x)}\) dx (or) ∫\(\sqrt{\frac{x-\alpha}{\beta-x}}\) dx  (or)∫\(\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\) dx	Put x = α cos2θ + β sin2θ Calculate (x – α)(β -x) and dx
6. ∫\(\frac{1}{a+b \sin x}\)dx (or) ∫\(\frac{1}{a+b \cos x}\)dx (or) ∫\(\frac{1}{a \sin x+b \cos x+c}\)dx (or)	Put t = tan\(\frac{x}{2}\) sin x = \(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\), cos x = \(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\)dx = \(\frac{2}{1+\tan ^2 \frac{x}{2}}\)dt
7. ∫\(\frac{1}{a+b \sin ^2 x}\)dx (or) ∫\(\frac{1}{a+b \cos ^2 x}\)dx (or) ∫\(\frac{1}{a \sin ^2 x+b \cos ^2 x+c \cos x \sin x}\)dx (or)	First multiply the numerator and denominator with sec2x and put t = tan x.
8. ∫\(\frac{a \cos x+b \sin x}{c \cos x+d \sin x}\)dx	Numerator = A (denominator) + B[\(\frac{d}{d x}\)(denominator)]
9. ∫\(\frac{a \sin x+b \cos x+c}{p \sin x+q \cos x+r}\)dx	Numerator = A (denominator) + B\(\frac{d}{d x}\)(denominator) + λ where A, B and λ are constants.

→ IntegratIon by parts: If U and V are two functions of x, then
∫UV dx = U∫V dx – ∫[\(\frac{dU}{d x}\)∫Vdx]dx

Extension Rule: ∫UVdx = UV₁ – U’V₂ + U”V₃ – U”’V₄ + ……….. + U where U’, U”, U”’ etc., are the successive derivatives of U and V₁, V₂, V₃ etc., are successive integrals of V.
Note : This rule is very useful if one of the Integrand Is an algebraic function.

→ Formulae:

• ∫e^ax sinbx dx = \(\frac{e^{a x}}{a^2+b^2}\)[a sin bx – b cos bx] + c
• ∫e^ax cos bx dx = \(\frac{e^{a x}}{a^2+b^2}\)[a cos bx – b sin bx] + c
• ∫e^x[f(x) + f'(x)]dx = e^xf(x) + c
• ∫e^-x[f(x) + f'(x)]dx = -e^-xf(x) + c
• ∫e^ax[af(x) + f'(x)]dx = e^axf(x) + c

→ Reduction Formulae:

• If I_n = ∫xⁿe^ax dx then I_n = \(\frac{x^n e^{a x}}{a}-\frac{n}{a}\)I_n-1
• If I_n = ∫sinⁿx dx then I_n = \(\frac{-\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n}\)I_n-2
• If I_n = ∫cosⁿxdx then I_n = \(\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n}\)I_n-2
• If I_n = ∫tanⁿx dx then I_n = \(\frac{\tan ^{n-1} x}{n-1}\) – I_n-2
• If I_n = ∫cotⁿx dx then I_n = \(\frac{-\cot ^{n-1} x}{n-1}\)I_n-2
• If I_n = ∫secⁿx dx then I_n = \(\frac{\sec ^{n-2} x \tan x}{n-1}+\frac{n-2}{n-1}\)I_n-2
• If I_n = ∫cosecⁿx dx then I_n = \(\frac{-{cosec}^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1}\)I_n-2
• If I_n = ∫(log x)ⁿ dx then I_n = x(log x)ⁿ – nI_n-1