Home » Ask & Discuss » Mathematics. » Integral Calculus « Back to Discussion

Integral Calculus

Blazing goIITian

Joined:	13 Aug 2008
Post:	1313

26 Mar 2009 13:06:19 IST

0 People liked this

456

!!! KILLER INTEGRATIONS !!!

None

Question 1 -

Question 2 -

Question 3 -

Question 4 -

Share this article on:

Comments (6)

spyduck

Hot goIITian

Joined: 28 Feb 2009

Posts: 119

26 Mar 2009 13:27:48 IST

Like 0 people liked this

those were my examples i did not number them sorry.

spyduck

Hot goIITian

Joined: 28 Feb 2009

Posts: 119

26 Mar 2009 13:30:35 IST

Like 0 people liked this

want more here u go

Forms containing Trigonometric Functions

$\int \sin u du = -\cos u + C$

$\int \cos u du = \sin u + C$

$\int \tan u du = \ln |\sec u| + C$

$\int \cot u du = \ln |\sin u| + C$

$\int \sec u du = ln |sec u + \tan u| + C = \ln |\tan \left (\frac {1}{4} \pi + \frac {1}{2} u \right)| + C$

$\int \csc u du = \ln |\csc u - \cot u| + C = \ln |\tan \frac {1}{2} u | + C$

$\int \sec^2 u du = \tan u + C$

$\int \csc^2 u du = -cot u + C$

$\int \sec u \tan u du = \sec u + C$

$\int \csc u \cot u du = -\csc u + C$

]Forms containing Inverse Trigonometric Functions

$\int \operatorname{sin^{-1}} \,u du = u \operatorname{sin^{-1}} \,u + \sqrt{1-u^2} + C$

$\int \operatorname{cos^{-1}} \,u du = u \operatorname{cos^{-1}} \,u - \sqrt{1-u^2} + C$

$\int \operatorname{tan^{-1}} \,u du = u \operatorname{tan^{-1}} \,u - ln{(1+u^2)} + C$

$\int \operatorname{cot^{-1}} \,u du = u \operatorname{cot^{-1}} \,u + ln{(1+u^2)} + C$

$\int \operatorname{sec^{-1}} \,u du = u \operatorname{sec^{-1}} \,u - ln | \,u + \sqrt{u^2 - 1} | + C = u \operatorname{sec^{-1}} \,u - \operatorname{cosh^{-1}} \,u + C$

$\int \operatorname{csc^{-1}} \,u du = u \operatorname{csc^{-1}} \,u + ln | \,u + \sqrt{u^2 - 1} | + C = u \operatorname{csc^{-1}} \,u - \operatorname{cosh^{-1}} \,u + C$

]Forms containing Exponential and Logarithmic Functions

$\int {e^u} du = e^u + C$

$\int {a^u} du = \frac{a^u}{\ln a} + C$

$\int {ue^u} du = e^u (u-1) + C$

$\int {u^n e^u} du = u^n e^u - n \int u^{n-1} e^u du$

$\int {u^n a^u} du = \frac {u^n a^u}{\ln a} - \frac {n}{\ln a} \int u^{n-1} a^u du + C$

$\int \frac {e^u}{u^n} du = -\frac{e^u}{(n-1)u^{n-1}} + \frac {1}{n-1} \int \frac {e^u}{u^{n-1}} du$

$\int \frac {a^u}{u^n} du = - \frac{a^u}{(n-1)u^{n-1}} + \frac {\ln a}{n-1} \int {a^u}{u^{n-1}} du$

$\int \ln u \,du = u \ln u - u + C$

$\int u^n \ln u \,du = \frac {u^{n+1}}{(n+1)^2} [(n+1)\ln {u - 1}] + C$

$\int \frac{1}{u \ln u} \,du = \ln | \ln u | + C$

$\int e^{au} \sin {nu} \,du = \frac {e^{au}}{a^2 + n^2} (a \sin nu - n \cos nu) + C$

$\int e^{au} \cos {nu} \,du = \frac {e^{au}}{a^2 + n^2} (a \cos nu + n \sin nu) + C$

]Forms containing Hyperbolic Functions

$\int\sinh ax\,dx = \frac{1}{a}\cosh ax + C$

$\int\cosh ax\,dx = \frac{1}{a}\sinh ax + C$

$\int \tanh ax\,dx = \frac{1}{a}\ln(\cosh ax) + C$

$\int \coth ax\,dx = \frac{1}{a}\ln(\sinh ax) + C$

$\int{\frac{du}{\sqrt{a^{2}+u^{2}}}}=\sinh ^{-1}\left( \frac{u}{a} \right)+C$

$\int{\frac{du}{\sqrt{u^{2}-a^{2}}}}=\cosh ^{-1}\left( \frac{u}{a} \right)+C$

$\int{\frac{du}{a^{2}-u^{2}}}=\frac{1}{a}\tanh ^{-1}\left( \frac{u}{a} \right)+C; u^{2}<a^{2}$

$\int{\frac{du}{a^{2}-u^{2}}}=\frac{1}{a}\coth ^{-1}\left( \frac{u}{a} \right)+C; u^{2}>a^{2}$

$\int{\frac{du}{u\sqrt{a^{2}-u^{2}}}}=-\frac{1}{a}\operatorname{sech}^{-1}\left( \frac{u}{a} \right)+C$

$\int{\frac{du}{u\sqrt{a^{2}+u^{2}}}}=-\frac{1}{a}\operatorname{csch}^{-1}\left| \frac{u}{a} \right|+C$

spyduck

Hot goIITian

Joined: 28 Feb 2009

Posts: 119

26 Mar 2009 13:39:32 IST

Like 0 people liked this

more more more

Properties of integration

Linearity

The collection of Riemann integrable functions on a closed interval [a, b] forms a vector space under the operations of pointwise addition and multiplication by a scalar, and the operation of integration

$f \mapsto \int_a^b f \; dx$

is a linear functional on this vector space. Thus, firstly, the collection of integrable functions is closed under taking linear combinations; and, secondly, the integral of a linear combination is the linear combination of the integrals,

$\int_a^b (\alpha f + \beta g)(x) \, dx = \alpha \int_a^b f(x) \,dx + \beta \int_a^b g(x) \, dx. \,$

Similarly, the set of real-valued Lebesgue integrable functions on a given measure space E with measure μ is closed under taking linear combinations and hence form a vector space, and the Lebesgue integral

$f\mapsto \int_E f d\mu$

is a linear functional on this vector space, so that

$\int_E (\alpha f + \beta g) \, d\mu = \alpha \int_E f \, d\mu + \beta \int_E g \, d\mu.$

More generally, consider the vector space of all measurable functions on a measure space (E,μ), taking values in a locally compact complete topological vector space V over a locally compact topological field K, f : E → V. Then one may define an abstract integration map assigning to each function f an element of V or the symbol ∞,

$f\mapsto\int_E f d\mu, \,$

that is compatible with linear combinations. In this situation the linearity holds for the subspace of functions whose integral is an element ofV (i.e. "finite"). The most important special cases arise when K is R, C, or a finite extension of the field Q_p of p-adic numbers, and V is a finite-dimensional vector space over K, and when K=C and V is a complex Hilbert space.

Linearity, together with some natural continuity properties and normalisation for a certain class of "simple" functions, may be used to give an alternative definition of the integral. This is the approach of Daniell for the case of real-valued functions on a set X, generalized by Nicolas Bourbaki to functions with values in a locally compact topological vector space. See (Hildebrandt 1953) for an axiomatic characterisation of the integral.

Inequalities for integrals

A number of general inequalities hold for Riemann-integrable functions defined on a closed and bounded interval [a, b] and can be generalized to other notions of integral (Lebesgue and Daniell).

Upper and lower bounds. An integrable function f on [a, b], is necessarily bounded on that interval. Thus there are real numbers m and M so that m ≤ f (x) ≤ M for all x in [a, b]. Since the lower and upper sums of f over [a, b] are therefore bounded by, respectively, m(b − a) and M(b− a), it follows that

$m(b - a) \leq \int_a^b f(x) \, dx \leq M(b - a).$

Inequalities between functions. If f(x) ≤ g(x) for each x in [a, b] then each of the upper and lower sums of f is bounded above by the upper and lower sums, respectively, of g. Thus

$\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx.$

This is a generalization of the above inequalities, as M(b − a) is the integral of the constant function with value M over [a, b].

Subintervals. If [c, d] is a subinterval of [a, b] and f(x) is non-negative for all x, then

$\int_c^d f(x) \, dx \leq \int_a^b f(x) \, dx.$

Products and absolute values of functions. If f and g are two functions then we may consider their pointwise products and powers, andabsolute values:

$(fg)(x)= f(x) g(x), \; f^2 (x) = (f(x))^2, \; |f| (x) = |f(x)|.\,$

If f is Riemann-integrable on [a, b] then the same is true for |f|, and

$\left| \int_a^b f(x) \, dx \right| \leq \int_a^b | f(x) | \, dx.$

Moreover, if f and g are both Riemann-integrable then f ², g ², and fg are also Riemann-integrable, and

$\left( \int_a^b (fg)(x) \, dx \right)^2 \leq \left( \int_a^b f(x)^2 \, dx \right) \left( \int_a^b g(x)^2 \, dx \right).$

This inequality, known as the Cauchy–Schwarz inequality, plays a prominent role in Hilbert space theory, where the left hand side is interpreted as the inner product of two square-integrable functions f and g on the interval [a, b].

Hölder's inequality. Suppose that p and q are two real numbers, 1 ≤ p, q ≤ ∞ with 1/p + 1/q = 1, and f and g are two Riemann-integrable functions. Then the functions |f|^p and |g|^q are also integrable and the following Hölder's inequality holds:

$\left|\int f(x)g(x)\,dx\right| \leq\left(\int \left|f(x)\right|^p\,dx \right)^{1/p} \left(\int\left|g(x)\right|^q\,dx\right)^{1/q}.$

For p = q = 2, Hölder's inequality becomes the Cauchy–Schwarz inequality.

Minkowski inequality. Suppose that p ≥ 1 is a real number and f and g are Riemann-integrable functions. Then |f|^p, |g|^p and |f + g|^p are also Riemann integrable and the following Minkowski inequality holds:

$\left(\int \left|f(x)+g(x)\right|^p\,dx \right)^{1/p} \leq\left(\int \left|f(x)\right|^p\,dx \right)^{1/p} +\left(\int \left|g(x)\right|^p\,dx \right)^{1/p}.$

An analogue of this inequality for Lebesgue integral is used in construction of L^p spaces.

Conventions

In this section f is a real-valued Riemann-integrable function. The integral

$\int_a^b f(x) \, dx$

over an interval [a, b] is defined if a < b. This means that the upper and lower sums of the function f are evaluated on a partition a = x₀ ≤ x₁ ≤ . . . ≤ x_n = b whose values x_i are increasing. Geometrically, this signifies that integration takes place "left to right", evaluating f within intervals [x_i ,x_i +1] where an interval with a higher index lies to the right of one with a lower index. The values a and b, the end-points of the interval, are called the limits of integration of f. Integrals can also be defined if a > b:

Reversing limits of integration. If a > b then define

$\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx.$

This, with a = b, implies:

Integrals over intervals of length zero. If a is a real number then

$\int_a^a f(x) \, dx = 0.$

The first convention is necessary in consideration of taking integrals over subintervals of [a, b]; the second says that an integral taken over a degenerate interval, or a point, should be zero. One reason for the first convention is that the integrability of f on an interval [a, b] implies that fis integrable on any subinterval [c, d], but in particular integrals have the property that:

Additivity of integration on intervals. If c is any element of [a, b], then

$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx.$

With the first convention the resulting relation

$\begin{align}\int_a^c f(x) \, dx &{}= \int_a^b f(x) \, dx - \int_c^b f(x) \, dx \\&{} = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx\end{align}$

is then well-defined for any cyclic permutation of a, b, and c.

Instead of viewing the above as conventions, one can also adopt the point of view that integration is performed on oriented manifolds only. If M is such an oriented m-dimensional manifold, and M' is the same manifold with opposed orientation and ω is an m-form, then one has (see below for integration of differential forms):

$\int_M \omega = - \int_{M'} \omega \,.$

Fundamental theorem of calculus

Main article: Fundamental theorem of calculus

The fundamental theorem of calculus is the statement that differentiation and integration are inverse operations: if a continuous function is first integrated and then differentiated, the original function is retrieved. An important consequence, sometimes called the second fundamental theorem of calculus, allows one to compute integrals by using an antiderivative of the function to be integrated.

Statements of theorems

Fundamental theorem of calculus. Let f be a real-valued integrable function defined on a closed interval [a, b]. If F is defined for x in [a, b] by

$F(x) = \int_a^x f(t)\, dt.$

then F is continuous on [a, b]. If f is continuous at x in [a, b], then F is differentiable at x, and F ′(x) = f(x).

Second fundamental theorem of calculus. Let f be a real-valued integrable function defined on a closed interval [a, b]. If F is a function such that F ′(x) = f(x) for all x in [a, b] (that is, F is an antiderivative of f), then

$\int_a^b f(t)\, dt = F(b) - F(a).$

Corollary. If f is a continuous function on [a, b], then f is integrable on [a, b], and F, defined by

$F(x) = \int_a^x f(t) \, dt$

is an anti-derivative of f on [a, b]. Moreover,

$\int_a^b f(t) \, dt = F(b) - F(a).$

Methods

Computing integrals

The most basic technique for computing definite integrals of one real variable is based on the fundamental theorem of calculus. It proceeds like this:

Let f(x) be the function of x to be integrated over a given interval [a, b].
Find an antiderivative of f, that is, a function F such that F' = f on the interval.
Then, by the fundamental theorem of calculus, provided the integrand and integral have no singularities on the path of integration,
$\int_a^b f(x)\,dx = F(b)-F(a).$

Note that the integral is not actually the antiderivative, but the fundamental theorem allows us to use antiderivatives to evaluate definite integrals.

The difficult step is often finding an antiderivative of f. It is rarely possible to glance at a function and write down its antiderivative. More often, it is necessary to use one of the many techniques that have been developed to evaluate integrals. Most of these techniques rewrite one integral as a different one which is hopefully more tractable. Techniques include:

Integration by substitution
Integration by parts
Changing the order of integration
Integration by trigonometric substitution
Integration by partial fractions
Integration by reduction formulae
Integration using parametric derivatives
Integrating trigonometric products as complex exponentials
Differentiation under the integral sign
Contour Integration

Properties of integration

Linearity

The collection of Riemann integrable functions on a closed interval [a, b] forms a vector space under the operations of pointwise addition and multiplication by a scalar, and the operation of integration

$f \mapsto \int_a^b f \; dx$

$\int_a^b (\alpha f + \beta g)(x) \, dx = \alpha \int_a^b f(x) \,dx + \beta \int_a^b g(x) \, dx. \,$

Similarly, the set of real-valued Lebesgue integrable functions on a given measure space E with measure μ is closed under taking linear combinations and hence form a vector space, and the Lebesgue integral

$f\mapsto \int_E f d\mu$

is a linear functional on this vector space, so that

$\int_E (\alpha f + \beta g) \, d\mu = \alpha \int_E f \, d\mu + \beta \int_E g \, d\mu.$

More generally, consider the vector space of all measurable functions on a measure space (E,μ), taking values in a locally compact complete topological vector space V over a locally compact topological field K, f : E → V. Then one may define an abstract integration map assigning to each function f an element of V or the symbol ∞,

$f\mapsto\int_E f d\mu, \,$

Inequalities for integrals

A number of general inequalities hold for Riemann-integrable functions defined on a closed and bounded interval [a, b] and can be generalized to other notions of integral (Lebesgue and Daniell).

Upper and lower bounds. An integrable function f on [a, b], is necessarily bounded on that interval. Thus there are real numbers m and M so that m ≤ f (x) ≤ M for all x in [a, b]. Since the lower and upper sums of f over [a, b] are therefore bounded by, respectively, m(b − a) and M(b− a), it follows that

$m(b - a) \leq \int_a^b f(x) \, dx \leq M(b - a).$

Inequalities between functions. If f(x) ≤ g(x) for each x in [a, b] then each of the upper and lower sums of f is bounded above by the upper and lower sums, respectively, of g. Thus

$\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx.$

This is a generalization of the above inequalities, as M(b − a) is the integral of the constant function with value M over [a, b].

Subintervals. If [c, d] is a subinterval of [a, b] and f(x) is non-negative for all x, then

$\int_c^d f(x) \, dx \leq \int_a^b f(x) \, dx.$

Products and absolute values of functions. If f and g are two functions then we may consider their pointwise products and powers, andabsolute values:

$(fg)(x)= f(x) g(x), \; f^2 (x) = (f(x))^2, \; |f| (x) = |f(x)|.\,$

If f is Riemann-integrable on [a, b] then the same is true for |f|, and

$\left| \int_a^b f(x) \, dx \right| \leq \int_a^b | f(x) | \, dx.$

Moreover, if f and g are both Riemann-integrable then f ², g ², and fg are also Riemann-integrable, and

$\left( \int_a^b (fg)(x) \, dx \right)^2 \leq \left( \int_a^b f(x)^2 \, dx \right) \left( \int_a^b g(x)^2 \, dx \right).$

Hölder's inequality. Suppose that p and q are two real numbers, 1 ≤ p, q ≤ ∞ with 1/p + 1/q = 1, and f and g are two Riemann-integrable functions. Then the functions |f|^p and |g|^q are also integrable and the following Hölder's inequality holds:

$\left|\int f(x)g(x)\,dx\right| \leq\left(\int \left|f(x)\right|^p\,dx \right)^{1/p} \left(\int\left|g(x)\right|^q\,dx\right)^{1/q}.$

For p = q = 2, Hölder's inequality becomes the Cauchy–Schwarz inequality.

Minkowski inequality. Suppose that p ≥ 1 is a real number and f and g are Riemann-integrable functions. Then |f|^p, |g|^p and |f + g|^p are also Riemann integrable and the following Minkowski inequality holds:

$\left(\int \left|f(x)+g(x)\right|^p\,dx \right)^{1/p} \leq\left(\int \left|f(x)\right|^p\,dx \right)^{1/p} +\left(\int \left|g(x)\right|^p\,dx \right)^{1/p}.$

An analogue of this inequality for Lebesgue integral is used in construction of L^p spaces.

Conventions

In this section f is a real-valued Riemann-integrable function. The integral

$\int_a^b f(x) \, dx$

Reversing limits of integration. If a > b then define

$\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx.$

This, with a = b, implies:

Integrals over intervals of length zero. If a is a real number then

$\int_a^a f(x) \, dx = 0.$

Additivity of integration on intervals. If c is any element of [a, b], then

$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx.$

With the first convention the resulting relation

$\begin{align}\int_a^c f(x) \, dx &{}= \int_a^b f(x) \, dx - \int_c^b f(x) \, dx \\&{} = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx\end{align}$

is then well-defined for any cyclic permutation of a, b, and c.

$\int_M \omega = - \int_{M'} \omega \,.$

Mirka

Blazing goIITian

Joined: 13 Aug 2008

Posts: 1313

26 Mar 2009 13:42:42 IST

Like 2 people liked this

Ahem, Mr. SpyDuck .......

but the Question was to " SOLVE THE GIVEN INTEGRALS "

nd not copy these stuff from Wiki

LOL

Anybody got an answer to the Questions on top ???

kabi

Blazing goIITian

Joined: 11 Jan 2009

Posts: 575

26 Mar 2009 14:58:49 IST

Like 1 people liked this

hey Mirka

i m starting from last as i got its answer only till now. I m trying others too.

integration ( x dx / ( x²-1)^3/2 )

put x² = t

so this integration will become integration ( dt / t^3/2 )

answer becomes = - 1 / (x² - 1)^1/2

now applying integration by parts in our original question with lnx as first func.

= - ln x / Sqrt (x² - 1) + integration ( dx/ ( x (x² -1)^1/2 ))

for second integration put x²-1 =t²

It will become integration ( dt / (t² + 1)) = tan^-1( Sqrt ( x² -1))

so ur answer

= tan^-1( (x²-1)^1/2 ) -lnx / (x² - 1)^1/2 + C

Thank u

I m trying others question too. So kindly wait.

kabi

Blazing goIITian

Joined: 11 Jan 2009

Posts: 575

26 Mar 2009 16:20:35 IST

Like 1 people liked this

Yo Got one more . But this method is quite big.

integration ( Sqrt ( sin(x) / ( 2 sinx + 3cosx) ) dx

= integration ( dx / Sqrt ( 2 + 3 cotx ))

put 2 + 3cotx = t²

- 3 cosec²x dx = 2 t dt

- 3 ( 1 + cot ² x ) dx = - 3 ( 1 + ( t ² - 2 ) ² / 9 ) dx = ( t⁴ + 13 - 4 t² )/ 3 dx = - 2t dt

dx = - 6 t dt / ( t⁴ + 13 - 4 t² )

integration ( - 6 dt / ( t⁴ + 13 - 4 t² ) )

= - 6 / ( 2 Sqrt (13 )) integration ( Sqrt (13 ) + t² + Sqrt (13) -t² ) dt / ( t⁴ + 13 - 4 t² ) )

= - 6 / ( 2 Sqrt (13 )) [ integration ( Sqrt (13 ) + t² )/ ( t⁴ + 13 - 4 t² )+ integration ( Sqrt ( 13) - t² ) /( t⁴ + 13 - 4 t² )

now take t² common from num and deno and write deno in the form ( t + Sqrt (13)/ t ) and ( t - Sqrt (13)/ t )

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team