Definitions [12]
The collection of all anti-derivatives of a function is called its indefinite integral.
Here: f(x) = integrand, dx = variable of integration, C = constant of integration.
A definite integral is connected with finding the area under a curve over a given interval. The chapter introduction presents area as one of the central motivating ideas behind integral calculus.
If the derivative of a function F(x) is f(x), then F(x) is called an antiderivative or integral of f(x). The set of all such antiderivatives is written as:
where C is an arbitrary constant called the constant of integration.
Integration by substitution is a method in which we replace a part of the integral by a new variable to simplify the integration.
General Formula:
If \[x = g(t), \ dx = g'(t) dt\] then \[\int f(x) dx = \int f(g(t))g'(t) dt\]
Integration using trigonometric identities means converting a trigonometric expression into an easier form with the help of standard identities before integrating.
Integration by partial fractions is a method used to integrate rational functions, that is, functions of the form
If two functions are written in the form uu and dvdv, then integration by parts is based on the product rule of differentiation.
\[\int\left(\mathrm{u.v}\right)\mathrm{dx}=\mathrm{u}\int\mathrm{v}\mathrm{dx}-\int\left(\frac{\mathrm{du}}{\mathrm{dx}}\right).\left(\int\mathrm{v}\mathrm{dx}\right)\mathrm{dx}\]
A definite integral represents the value of a function accumulated between two limits.
It can also be interpreted geometrically as the net area between the graph of y = f(x) and the x-axis from x = a to x = b.
If a function f is continuous on an interval, the area function is defined by
This means that A(x) gives the area accumulated from x = a to a variable point x.

If f(x) is a continuous function defined on an interval [a, b] and if Φ(x) is the antiderivative of f(x), i.e., \[\frac{d}{dx}[\phi(x)]=f(x)\] then the definite integral of f(x) over [a, b] denoted by \[\int_{a}^{b}f(x)dx\] is defined as
\[\int_{a}^{b}f(x)dx=
\begin{bmatrix}
\phi\left(x\right)
\end{bmatrix}_{a}^{b}=\phi\left(b\right)-\phi\left(a\right)\]
\[\mathrm{If~}\frac{d}{dx}[F(x)]=f(x),\mathrm{~then~}\int f(x)dx=F(x)\]
Integration is the inverse process of differentiation.
\[\int f(x)dx=F(x)+c\]
- The arbitrary constant 'c' is called the constant of integration.
- F(x) + c is called the indefinite integral.
Formulae [12]
| No. | Derivatives | Integrals (Anti-derivatives) |
|---|---|---|
| (i) | \[\frac{d}{dx} \left( \frac{x^{n+1}}{n+1} \right) = x^n\]; | \[\int x^n dx = \frac{x^{n+1}}{n+1} + \text{C}, n \neq -1\] |
| \[\frac{d}{dx} (x) = 1\]; | \[\int dx = x + \text{C}\] | |
| (ii) | \[\frac{d}{dx} (\sin x) = \cos x\]; | \[\int \cos x dx = \sin x + \text{C}\] |
| (iii) | \[\frac{d}{dx} (-\cos x) = \sin x\]; | \[\int \sin x dx = -\cos x + \text{C}\] |
| (iv) | \[\frac{d}{dx} (\tan x) = \sec^2 x\]; | \[\int \sec^2 x dx = \tan x + \text{C}\] |
| (v) | \[\frac{d}{dx} (-\cot x) = \text{cosec}^2 x\]; | \[\int \text{cosec}^2 x dx = -\cot x + \text{C}\] |
| (vi) | \[\frac{d}{dx} (\sec x) = \sec x \tan x\]; | \[\int \sec x \tan x dx = \sec x + \text{C}\] |
| (vii) | \[\frac{d}{dx} (-\text{cosec} x) = \text{cosec} x \cot x\]; | \[\int \text{cosec} x \cot x dx = -\text{cosec} x + \text{C}\] |
| (viii) | \[\frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}\]; | \[\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + \text{C}\] |
| (ix) | \[\frac{d}{dx} (-\cos^{-1} x) = \frac{1}{\sqrt{1-x^2}}\]; | \[\int \frac{dx}{\sqrt{1-x^2}} = -\cos^{-1} x + \text{C}\] |
| (x) | \[\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}\]; | \[\int \frac{dx}{1+x^2} = \tan^{-1} x + \text{C}\] |
| (xi) | \[\frac{d}{dx} (e^x) = e^x\]; | \[\int e^x dx = e^x + \text{C}\] |
| (xii) | \[\frac{d}{dx}\left(\log|x|\right)=\frac{1}{x};\] | \[\int\frac{1}{x}dx=\log|x|+\mathrm{C}\] |
| (xiii) | \[\frac{d}{dx} \left( \frac{a^x}{\log a} \right) = a^x\]; | \[\int a^x dx = \frac{a^x}{\log a} + \text{C}\] |
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\[\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C\]
-
\[\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C\]
-
\[\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C\]
-
\[\int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C\]
-
\[\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C\]
-
\[\int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C\]
- \[\int[f(x)]^nf^{\prime}(x)dx=\frac{[f(x)]^{n+1}}{n+1}+c\quad(n\neq-1)\]
-
\[\int\frac{f^{\prime}(x)}{f(x)}dx=\log|f(x)|+c\]
| Expression | Equivalent Form |
|---|---|
| \[\sin^2x+\cos^2x\] | 1 |
| \[1+\tan^2x\] | \[sec^2x\] |
| \[1+\cot^2x\] | \[cosec^2x\] |
| \[\sin^2x\] | \[\frac{1-\cos2x}{2}\] |
| \[\cos^2x\] | \[\frac{1+\cos2x}{2}\] |
| sin x cos x | \[\frac{1}{2}\sin2x\] |
| sin x cos y | \[\frac{1}{2}[\sin(x+y)+\sin(x-y)]\] |
| cos x sin y | \[\frac{1}{2}[\sin(x+y)-\sin(x-y)]\] |
| cos x cos y | \[\frac{1}{2}[\cos(x+y)+\cos(x-y)]\] |
| sin x sin y | \[\frac{1}{2}[\cos(x-y)-\cos(x+y)]\] |
| 1 - cos x | \[2\sin^2\frac{x}{2}\] |
| 1 + cos x | \[2\cos^2\frac{x}{2}\] |
| \[\sin^3x\] | \[\frac{1}{4}(3\sin x-\sin3x)\] |
| \[cos^3x\] | \[\frac{1}{4}(3\cos x+\cos3x)\] |
| Function | Integral |
|---|---|
| \[\int\tan x\mathrm{~}dx\] | \[\log|\sec x|+c\] |
| \[\int\cot x\mathrm{~}dx\] | \[\log|\sin x|+c\] |
| \[\int\sec x\operatorname{d}x\] | \[\log|\sec x+\tan x|+c\] |
| \[\int cosecxdx\] | \[\log|\left(\csc x-\cot x\right)|+c\] |
\[\int e^x\left[\left.f(x)+f^{\prime}(x)\right.\right]dx=e^xf(x)+c\]
If, u = f(x) ⇒ \[\frac{du}{dx}=f^{\prime}(x)\]
then \[\int[f(x)]^nf^{\prime}(x)dx=\frac{[f(x)]^{n+1}}{n+1}+c\quad(n\neq-1)\]
Linear Substitution Rule:
If u = ax + bu = , then
\[\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c\quad(n\neq-1)\]
| No. | Differentiation | Integration |
|---|---|---|
| 1 | \[\frac{d}{dx}(x^{n+1})=(n+1)x^n\] | \[\int x^ndx=\frac{x^{n+1}}{n+1}+c\] |
| 2 | \[\frac{d}{dx}(\log x)=\frac{1}{x}\] | \[\int\frac{1}{x}dx=\log\mid x\mid+c\] |
| 3 | \[\frac{d}{dx}(e^x)=e^x\] | \[\int e^{x}dx=e^{x}+c\] |
| 4 | \[\frac{d}{dx}(a^x)=a^x\log_ea\] | \[\int a^{x}dx=\frac{a^{x}}{\log_{e}a}+c(a>0,a\neq1)\] |
| 5 | \[\frac{d}{dx}(\sin x)=\cos x\] | \[\int\cos xdx=\sin x+c\] |
| 6 | \[\frac{d}{dx}(\cos x)=-\sin x\] | \[\int\sin xdx=-\cos x+c\] |
| 7 | \[\frac{d}{dx}(\tan x)=\sec^2x\] | \[\int\sec^2xdx=\tan x+c\] |
| 8 |
\[\frac{d}{dx}(\cot x)=-\mathrm{cosec}^{2}x\] |
\[\int\mathrm{cosec}^2xdx=-\cot x+c\] |
| 9 | \[\frac{d}{dx}(\sec x)=\sec x\tan x\] | \[\int\sec x\tan xdx=\sec x+c\] |
| 10 | \[\frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x\] | \[\int\operatorname{cosec}x\cot xdx=-\operatorname{cosec}x+c\] |
| 11 |
\[\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}\] \[\frac{d}{dx}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^{2}}}\] |
\[\begin{aligned} & \int{\frac{1}{\sqrt{1-x^{2}}}}dx=\sin^{-1}x+c \\ \mathrm{OR} & =-\cos^{-1}x+c \end{aligned}\] |
| 12 |
\[\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\] \[\frac{d}{dx}(\cot^{-1}x)=\frac{-1}{1+x^{2}}\] |
\[\int\frac{1}{1+x^{2}}dx=\tan^{-1}x+c\mathrm{OR}=-\cot^{-1}x+c\] |
| 13 |
\[\frac{d}{dx}(\sec^{-1}x)=\frac{1}{x\sqrt{x^{2}-1}}\] \[\frac{d}{dx}(\mathrm{cosec}^{-1}x)=\frac{-1}{x\sqrt{x^{2}-1}}\] |
\[\int\frac{1}{x\sqrt{x^{2}-1}}dx=\sec^{-1}x+cOR=-cosec^{-1}x+c\] |
| 14 | \[\frac{d}{dx}\left(\sin^{-1}\frac{x}{a}\right)=\frac{1}{\sqrt{a^{2}-x^{2}}}\] | \[\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\frac{x}{a}+c\] |
| 15 | \[\frac{d}{dx}\left(\tan^{-1}\frac{x}{a}\right)=\frac{a}{a^2+x^2}\] | \[\int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}+c\] |
| 16 | \[\frac{d}{dx}\left(\sec^{-1}\frac{x}{a}\right)=\frac{a}{x\sqrt{x^{2}-a^{2}}}\] | \[\int\frac{dx}{x\sqrt{x^{2}-a^{2}}}dx=\frac{1}{a}\sec^{-1}\frac{x}{a}+c\] |
Statement:
If f(x) and g(x) are any two differentiable functions of x and G(x) is the antiderivative of g(x), i.e., \[G(x)=\int g(x)dx\]. Then
\[\int f(x)g(x)dx=f(x)G(x)-\int f^{\prime}(x)G(x)dx\]
| Integral | Result |
|---|---|
| \[\int\frac{dx}{x^2+a^2}\] | \[\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+c\] |
| \[\int\frac{dx}{x^2-a^2}\] | \[\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+c\] |
| \[\int\frac{dx}{a^2-x^2}\] | \[\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+c\] |
(A) Non-repeated linear factors
\[\frac{Ax+B}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}\]
(B) Repeated linear factor
\[\frac{Ax+B}{(x-a)^n}=\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n}\]
(C) Quadratic factor (not factorisable)
\[\frac{Ax+B}{ax^2+bx+c}\]
1.\[\int\sqrt{(a^{2}-x^{2})}dx=\frac{1}{2}x\sqrt{(a^{2}-x^{2})}+\frac{1}{2}a^{2}\sin^{-1}\left(\frac{x}{a}\right)+c\]
2. \[\int\left(\sqrt{a^{2}+x^{2}}\right)dx=\frac{1}{2}x\sqrt{(a^{2}+x^{2})}+\frac{1}{2}a^{2}\log|x+\sqrt{(a^{2}+x^{2})}|+c\]
Theorems and Laws [4]
Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`
Let I = `int sqrt(a^2 - x^2) dx`
= `int sqrt(a^2 - x^2)*1 dx`
= `sqrt(a^2 - x^2)* int 1 dx - int [d/dx (sqrt(a^2 - x^2))* int 1 dx]dx`
= `sqrt(a^2 - x^2)*x - int [1/(2sqrt(a^2 - x^2))*d/dx (a^2 - x^2)*x]dx`
= `sqrt(a^2 - x^2)*x - int 1/(2sqrt(a^2 - x^2))(0 - 2x)*x dx`
= `sqrt(a^2 - x^2)*x - int (-x)/sqrt(a^2 - x^2)*x dx`
= `xsqrt(a^2 - x^2) - int (a^2 - x^2 - a^2)/sqrt(a^2 - x^2)dx`
= `xsqrt(a^2 - x^2) - int sqrt(a^2 - x^2)dx + a^2 int dx/sqrt(a^2 - x^2)`
= `xsqrt(a^2 - x^2) - I + a^2sin^-1(x/a) + c_1`
∴ 2I = `xsqrt(a^2 - x^2) + a^2sin^-1(x/a) + c_1`
∴ I = `x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1(x/a) + c_1/2`
∴ `int sqrt(a^2 - x^2)dx = x/2 sqrt(a^2 - x^2) + a^2/2sin^-1(x/a) + c`, where `c = c_1/2`.
If f is continuous on [a, b] and
If f is continuous on [a, b] and F is any antiderivative of f, then
This is the formula most often used in exams to evaluate definite integrals.
Theorem 1:
Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function. Then A′(x) = f (x), for all x ∈ [a, b]
Theorem 2:
Let f be a continuous function defined on the closed interval [a, b], and F be an antiderivative of f. Then \[\int_a^bf(x)dx=\left[\mathbf{F}(x)\right]_a^b=\mathbf{F}(b)-\mathbf{F}(a)\]
Key Points
- Primitive
Another name for anti-derivative. - Indefinite Integral
The family of all anti-derivatives of a function. - Definite Integral
An integral taken between two fixed limits, commonly used to represent area or total accumulation. - Integral Calculus
The branch of calculus dealing with anti-derivatives, accumulation, and areas under curves.
-
Integration is the inverse process of differentiation.
-
The result of indefinite integration is called the antiderivative or primitive.
-
General form: ∫f(x) dx = F(x) + C.
-
The constant CC must always be added in indefinite integrals.
| Property | Formula |
| Reverse of differentiation | \[\frac{d}{dx}\left(\int f(x) dx\right) = f(x)\] |
| Same derivative | \[F'(x) = G'(x) \Rightarrow F(x) = G(x) + C\] |
| Sum rule | \[\int (f + g)dx = \int fdx + \int gdx\] |
| Difference rule | \[\int (f - g)dx = \int fdx - \int gdx\] |
| Constant multiple rule | \[\int k f(x)dx = k \int f(x)dx\] |
| General linearity | \[\int (kf \pm lg)dx = k \int fdx \pm l \int gdx\] |
-
Integration by substitution is the reverse process of the chain rule.
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Choose the substitution so that the integral becomes simpler, not more complicated.
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Always rewrite both the function and \(dx\) in terms of the new variable.
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For indefinite integrals, back-substitute and add \(C\).
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For definite integrals, limits should also be changed if the solution is continued in the new variable.
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Trigonometric substitution is mainly used for radicals involving \(a^2-x^2\), \(x^2+a^2\), and \(x^2-a^2\).
-
First inspect the pattern in the integrand.
-
Do not integrate complicated trigonometric expressions directly if an identity can simplify them first.
-
After simplification, integrate term by term carefully.
-
Always add the constant of integration, \(C\).
-
First check whether the rational function is proper or improper.
-
Use long division before decomposition if the fraction is improper.
-
Factorise the denominator completely before choosing partial fractions.
-
For each distinct linear factor, use a constant numerator such as A, B, or C.
-
For a repeated linear factor, include every power separately.
-
For an irreducible quadratic factor, use a linear numerator of the form Bx + C.
-
After decomposition, integrate each term separately using standard formulas.
-
Formula:
\[\int u dv = uv - \int v du\] -
Choose u by LIATE
-
For log x and inverse trig, multiply by 1
-
Repeated parts may be needed for \[e^x \sin x\], \[e^x \cos x\].
-
Convert the integrand into a known standard form before integrating.
-
For \[x^2 - a^2\], factorize and use partial fractions.
-
For \[x^2 + a^2\], the answer usually involves \[\tan^{-1}\].
-
For \[\sqrt{a^2 - x^2}\], the answer usually involves \[\sin^{-1}\].
-
For general quadratics, complete the square first.
-
For \[px + q\] in the numerator, relate it to the derivative of the denominator.
-
Always write the constant of integration C in the final answer.
-
Used to find exact accumulated value over a fixed interval.
-
Written as \[\int_{a}^{b} f(x) \, dx\].
-
Evaluated using \[F(b) - F(a)\].
-
Gives a unique numerical value.
-
Represents net area geometrically.
-
The theorem connects differentiation and integration.
-
If \[A(x) = \int_{a}^{x} f(t) \, dt\], then \[A'(x) = f(x)\].
-
If F'(x) = f(x), then \[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\].
-
The result is used to evaluate definite integrals quickly.
-
The function should be continuous on the interval for direct use of the theorem.
-
Look for an inner function and its derivative.
-
Choose substitution carefully.
-
Change the limits immediately.
-
Integrate in the new variable.
-
Do not add +C in a definite integral.
For choosing the first function:
L I A T E
-
Logarithmic
-
Inverse trigonometric
-
Algebraic
-
Trigonometric
-
Exponential
1.\[\frac{d}{dx}{\left[\int f(x)dx\right]}=f(x)\]
2. \[\int cf(x)dx=c\int f(x)dx\]
3. \[\int(u+v-w)dx=\int udx+\int vdx-\int wdx\]
Important Questions [131]
- Find : ∫ ( X 2 + 1 ) ( X 2 + 4 ) ( X 2 + 3 ) ( X 2 − 5 ) D X .
- Anti-derivative of tanx-1tanx+1 with respect to x is ______.
- If f(x) =∫x0 t sin t dt, then write the value of f '(x).
- int x^2 e^(x^3) dx equals
- Write the antiderivative of (3√x+1/√x).
- Evaluate : ∫(sin^6x+cos^6x)/(sin^2x.cos^2x)dx
- Find :∫(x2+x+1)/((x2+1)(x+2))dx
- Evaluate : π 4 ∫ 0 Tan X D X .
- If ddxf(x)=2x+3x and f(1) = 1, then f(x) is ______.
- Find Integral Dx by 5−8x− X Sqrt
- Prove that int_0^"a" "f(x)" "dx" = int_0^"a" "f"("a"-"x")"dx" ,and hence evaluate int_0^1 "x"^2(1 - "x")^"n""dx".
- Find Integral Cos Theta by 4 + Sin Square Thetax5 - 4 Cos Square Theta
- Evaluate `Int (Cos 2x + 2sin^2x)/(Cos^2x) Dx`
- Evaluate : π ∫ 0 X Tan X Sec X + Tan X D X .
- Find: ∫(x+3)3-4x-x2dx
- Evaluate : ∫(x+2)/√(x2+5x+6)dx
- Evaluate: ∫1/cos4x+sin4x dx
- Evaluate : ∫1/(sin^4x+sin^2xcos^2x+cos^4x)dx
- Find α∫dxsin3xcos(x-α).
- Evaluate ∫log2log31(ex+e-x)(ex-e-x)dx.
- Find ∫(3 sin 𝜃 − 2)cos𝜃/5 − cos^2𝜃 − 4 sin 𝜃 𝑑𝜃
- Find : ∫ Sin 2 X ( Sin 2 X + 1 ) ( Sin 2 X + 3 ) D X
- Find ∫x+2x2-4x-5dx.
- Find ∫√x/√(a^3−x^3)dx
- Find : int((2x-5)e(2x))/(2x-3)3dx
- Find the particular solution of the differential equation x2dy = (2xy + y2) dx, given that y = 1 when x = 1.
- Evaluate ∫-aaf(x)dx, where f(x) = 9x1+9x.
- ∫secxsecx-tanxdx equals ______.
- Evaluate : ∫(x−3)√(x2+3x−18x) dx
- Evaluate : ∫(√cotx+√tanx)dx
- Find ∫ Sin ( X − a ) Sin ( X + a ) D X
- Evaluate : ∫sin(x−a)/sin(x+a)dx
- Evaluate :∫π/3 π/6 dx/(1+√cotx)
- Find: ∫ Sin − 1 ( 2 X ) D X .
- Find: Intsqrt(1 - Sin 2x) Dx, Pi/4 < X < Pi/2
- Find: Int Sec^2 X /Sqrt(Tan^2 X+4) Dx.
- Integrate the function cos("x + a")/sin("x + b")w.r.t. x.
- Find ∫ Sin X − Cos X √ 1 + Sin 2 X D X , 0 < X < π 2
- Find: ∫ Cos X ( 1 + Sin X ) ( 2 + Sin X ) D X
- Find the Area of the Triangle Whose Vertices Are (-1, 1), (0, 5) and (3, 2), Using Integration.
- Find `Integral Dx/(X^2 + 4x + 8)`
- Find ∫ ( Log X ) 2 D X
- Evaluate: int_0^π (x sin x)/(1 + cos^2x) dx.
- Evaluate `3by2integral0|X Sin Pi X|Dx`
- Find `Integral(2x)By((Xsquare2 + 1)(Xpower4 + 4))`Dx
- Find Integral((3 Sin X - 2) Cos X)/(By3 - Cos^2 X- 7 Sin X) Dx
- Evaluate : π ∫ 0 X Tan X Sec X ⋅ C O S E C X D X .
- Find: int"dx"/sqrt(5-4"x" - 2"x"^2)
- Find Integral (Sin 2 X - Cos 2x)By(Sin X Cos X) Dx
- Find : ∫ Sin ( X − a ) Sin ( X + a ) D X
- Find: ∫x2(x2+1)(3x2+4)dx
- Find : ∫x2x4+x2−2dx
- Evaluate: ∫ x^2/(x^4+x^2-2)dx
- Find Integral(E^X Dx)By((E^X - 1)Square2 (Ex + 2))`
- Evaluate: ∫-215-4x-x2dx
- Find `Int (2cos X)/((1-sinx)(1+Sin^2 X)) Dx`
- Find: I=intdx/(sinx+sin2x)
- Find: ∫x4(x-1)(x2+1)dx.
- Evaluate: ∫0π4dx1+tanx
- Evaluate ∫π0 e^2 x.sin(π/4+x) dx
- Find : ∫ ( Log X ) 2 D X
- Find: ∫ex.sin2xdx
- Find: ∫2x(x2+1)(x2+2)dx
- Find the general solution of the differential equation: edydx=x2.
- Find ∫ecot-1x(1-x+x21+x2)dx.
- Find ∫sin-1x(1-x2)3/2dx.
- Find ∫ex(1-sinx1-cosx)dx.
- Find: ∫ex2(x5+2x3)dx.
- Find `Integral (2x)By(Xsquare2 + 1)(Xsquare2 + 2)Square2 Dx`
- Evaluate: ∫(x+3)e^x/(x+5)^3dx
- find : ∫(3x+1)√(4-3x-2x^2)dx
- Find ∫dx4x-x2
- Find Integral of Dx by (5 - 8x - X2)
- Evaluate 3integral2 3 Powerx Dx`
- Find: int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"
- Find : ∫ ( 2 X + 5 ) √ 10 − 4 X − 3 X 2 D X .
- Evaluate: ∫(5x-2)/(1+2x+3x^2)dx
- Evaluate : ∫x^2 (x^2+4) (x^2+9)dx
- Find: ∫(x^3−1)/(x^3+x) dx
- Find: ∫dxx2-6x+13
- Find : ∫ B a Log X X Dx
- Evaluate Each of the Following Integral: ∫ π 2 0 E X ( Sin X − Cos X ) D X
- Prove that ∫ B a ƒ ( X ) D X = ∫ B a ƒ ( a + B − X ) D X and Hence Evaluate ∫ π 3 π 6 D X 1 + √ Tan X
- Evaluate : π ∫ 0 / 4 Sin X + Cos X 16 + 9 Sin 2 X D X .
- Evaluate : 2 π ∫ 0 Cos 5 X D X .
- Evaluate : \[\Int E^{2x} \Cdot \Sin \Left( 3x + 1 \Right) Dx\] .
- Evaluate : ∫ D X Sin 2 X Cos 2 X .
- Evaluate: π / 2 ∫ − π / 2 Cos X 1 + E X D X .
- Evaluate : π ∫ 0 X 1 + Sin α Sin X D X
- Evaluate :∫π/2 0 2sinx/(2sinx+2cosx)dx
- Evaluate : ∫π/2 0 (sin^2 x)/(sinx+cosx)dx
- Evaluate : ∫ X Cos − 1 X √ 1 − X 2 D X .
- Find : ∫ E 2 X Sin ( 3 X + 1 ) D X .
- \[\Int\Limits_{- 2}^1 \Left| X^3 - X \Right|Dx\]
- Evaluate `Int_1^3 (X^2 + 3x + E^X) Dx` as the Limit of the Sum
- Evaluate: ∫ π − π ( 1 − X 2 ) Sin X Cos 2 X D X .
- Evaluate: ∫ 2 − 1 | X | X D X .
- Evaluate: ∫ 5 1 { | X − 1 | + | X − 2 | + | X − 3 | } D X .
- Find: ∫ ( 3 X + 5 ) √ 5 + 4 X − 2 X 2 D X .
- ∫e2e dx/(xlogx)
- If ∫a0 1/(4+x2)dx=π/8 , find the value of a.
- Evaluate : ∫π0 (4x sin x)/(1+cos2 x) dx
- find ∫42 x/(x2+1) dx
- ∫π−π (cos ax−sin bx)2 dx
- Evaluate ∫2−1 ∣x^3−x∣ dx
- Evaluate: π∫0π2sin2xtan-1(sinx)dx.
- Evaluate ∫0(3/2) |x cosπx| dx
- Evaluate : ∫40(|x|+|x−2|+|x−4|)dx
- Find : ∫ X Sin − 1 X √ 1 − X 2 D X .
- Evaluate: π / 2 ∫ 0 X Sin X Cos X Sin 4 X + Cos 4 X D X .
- Evaluate: ∫13xx+4-xdx
- Evaluate π∫0π/4log(1+tanx)dx.
- Evaluate: π∫0π211+(tanx)23dx
- The value of π∫0π4(sin2x)dx is ______.
- Prove that ∫ b a f ( x ) d x = ∫ b a f ( a + b − x ) d x and hence evaluate ∫ π 3 π 6 d x 1 + √ tan x .
- Evaluate ∫-11|x4-x|dx.
- Evaluate: ππ∫-π/4π/4cos2x1+cos2xdx.
- Evaluate: π∫0πx1+sinxdx.
- If ππ∫02πcos2x dx=k∫0π2cos2x dx, then the value of k is ______.
- Assertion (A): ∫2810-xx+10-xdx = 3. Reason (R): ∫abf(x)dx=∫abf(a+b-x)dx.
- Evaluate the definite integrals ∫0πxtanxsecx+tanxdx
- Evaluate : ∫ ( 3 X − 2 ) √ X 2 + X + 1 D X .
- Evaluate: 1integral4 {|X -1|+|X - 2|+|X - 4|}Dx`
- Find : ∫ 2 X + 1 ( X 2 + 1 ) ( X 2 + 4 ) D X .
- ∫-11|x-2|x-2dx, x ≠ 2 is equal to ______.
- Evaluate: ∫ π 0 X Sin X 1 + 3 Cos 2 X D X .
- Prove that ∫ a 0 F ( X ) D X = ∫ a 0 F ( a − X ) D X , Hence Evaluate ∫ π 0 X Sin X 1 + Cos 2 X D X
- Evaluate: π∫02π11+esinxdx
- Evaluate ∫2−2 x2/(1+5x) dx
- Evaluate: ∫-13|x3-x|dx
- Evaluate: ππ∫-π2π2(sin|x|+cos|x|)dx
Concepts [13]
- Introduction of Integrals
- Integration as an Inverse Process of Differentiation
- Properties of Indefinite Integral
- Methods of Integration> Integration by Substitution
- Methods of Integration>Integration Using Trigonometric Identities
- Methods of Integration> Integration Using Partial Fraction
- Methods of Integration> Integration by Parts
- Integrals of Some Particular Functions
- Definite Integrals
- Fundamental Theorem of Integral Calculus
- Evaluation of Definite Integrals by Substitution
- Properties of Definite Integrals
- Overview of Integrals
