how do we integrate this expression f(x)=sin(1/4)x? I know how do differentiate it however I was confused how to integrate it.

i would suppose you meant sin(x/4) and not xsin(1/4) so to integrate that you can concider x/4=t so we have sin(x/4)dx sin(t)dx now as x/4=t differentiate wrt xto get 1/4=dt/dx dx=4dt so we have sin(t)dx=sin(t)4dt 4sin(t)dt=c - 4cos(t) t=x/4 therefore we have c - 4cos(x/4) thus sin(x/4)dx=c - 4cos(x/4) little tip/trick:: integrate normally concidering t then divide by the derivative(should be a constant) eg:-sin(x/4)dx --> integral of sint is -cost and t is x/4 so its derivative is 1/4 thus the answer will be (-cos(x/4))/(1/4)= -4cos(x/4) [we divided by 1/4 and not 4] [dont forget the c]

How should I integrate 1/x? They say the answer is ln(x) but doesn't it make sense that the answer would be zero?

That would mean that the area under the graph of 1/x is 0, which is clearly not the case. To integrate 1/x, you need to understand how to differentiate ln(x).

how do we integrate csc^2 (x)*sec^2 (x)?

Let, I = ∫csc2x.sec2x.dx = ∫ (1/ sin〗^2⁡x ).(1/〖cos〗^2⁡x ).dx = 4 ∫ 1/〖4 sin〗^2⁡〖x .〖cos〗^2⁡x 〗 dx = 4 ∫ 1/(2sinx.cosx)^2 .dx = 4 ∫ 1/(sin2x)^2 dx = 4 ∫ csc22x.dx = 4 ((-cot2x))/2 + C = - 2cot2x + C = - 2 ((1-〖tan〗^2⁡x)/2tanx) + C = - ((1-〖tan〗^2⁡〖x)〗)/tanx + C = (〖tan〗^2⁡x - 1)/tanx + C I = tanx – cotx + C

Main content

Class 12 math (India)

Course: Class 12 math (India) > Unit 9

Lesson 3: Indefinite integrals of common functions

Antiderivatives and indefinite integrals review

Google Classroom

Review your knowledge of antiderivatives and indefinite integrals.

What are antiderivatives and indefinite integrals?

Antiderivative is the reverse relation of derivative. For example, we know that the derivative of

x^{2}

2 x

. This means that an antiderivative of

2 x

x^{2}

$f^{'}$ ‍ is the derivative of $f$ ‍ $⟺$ ‍ $f$ ‍ is an antiderivative of $f^{'}$ ‍

Each function has a family of antiderivatives. For example, the antiderivatives of

2 x

are the family of functions

x^{2} + c

where

c

can be any constant number.

The indefinite integral of a function can be viewed as exactly that, the family of antiderivatives of the function. It also has a special notation. For example, the indefinite integral of

2 x

is expressed as

\int 2 x d x

In general,

\int f^{'} (x) d x = f (x) + c

Want to learn more about antiderivatives and indefinite integrals? Check out this video.

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emzace17
Posted 7 years ago. Direct link to emzace17's post “how do we integrate this ...”
how do we integrate this expression f(x)=sin(1/4)x? I know how do differentiate it however I was confused how to integrate it.
Button navigates to signup pageComment on emzace17's post “how do we integrate this ...”
(1 vote)
Answer
- vaibhav
  Posted 7 years ago. Direct link to vaibhav's post “i would suppose you meant...”
  i would suppose you meant sin(x/4) and not xsin(1/4)
  so to integrate that you can concider x/4=t
  so we have
  sin(x/4)dx
  sin(t)dx
  now as x/4=t
  differentiate wrt xto get
  1/4=dt/dx
  dx=4dt
  so we have
  sin(t)dx=sin(t)4dt
  4sin(t)dt=c - 4cos(t)
  t=x/4
  therefore we have
  c - 4cos(x/4)
  thus sin(x/4)dx=c - 4cos(x/4)
  
  little tip/trick:: integrate normally concidering t then divide by the derivative(should be a constant)
  eg:-sin(x/4)dx --> integral of sint is -cost and t is x/4 so its derivative is 1/4 thus the answer will be (-cos(x/4))/(1/4)= -4cos(x/4) [we divided by 1/4 and not 4] [dont forget the c]
  Button navigates to signup page
  (27 votes)
Jonathan Wulff
Posted 6 years ago. Direct link to Jonathan Wulff's post “How should I integrate 1/...”
How should I integrate 1/x? They say the answer is ln(x) but doesn't it make sense that the answer would be zero?
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(1 vote)
Answer
- kubleeka
  Posted 6 years ago. Direct link to kubleeka's post “That would mean that the ...”
  That would mean that the area under the graph of 1/x is 0, which is clearly not the case. To integrate 1/x, you need to understand how to differentiate ln(x).
  Button navigates to signup page
  (4 votes)
Mr. Oria
Posted 7 years ago. Direct link to Mr. Oria's post “How should we integrate s...”
How should we integrate sin sqrt(x) using reverse chain rule?
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(2 votes)
Answer
- Arunabh Bhattacharya
  Posted 7 years ago. Direct link to Arunabh Bhattacharya's post “let u = x ^ 1/2, then du/...”
  let u = x ^ 1/2, then du/dx = (1/2) * x ^ -1/2
  dx = 2 x ^ 1/2 du or dx = 2 u du
  ∫ sin (x ^ 1/2) dx = ∫ sin (u) * 2 u du
  Let f'(u) = sin (u) and g(u) = 2u, then g'(u) = 2 and f(u) = - cos (u)
  Applying reverse product rule:
  ∫ f ʹ(u) * g(u) du = f(u) * g(u) - ∫ f(u) * gʹ(u) du
  ∫ 2u * sin (u) du = 2 u * (- cos (u)) - ∫ 2 * (- cos (u)) du
  = 2 u * (- cos (u)) - 2 (- sin (u)) + C
  Substituting u = x^ 1/2
  ∫ sin (x ^ 1/2) dx = 2 sin (x ^ 1 / 2) - 2 x ^ 1/2 * cos (x ^ 1/2) + C
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  (0 votes)
1151104828
Posted 7 years ago. Direct link to 1151104828's post “how do we integrate csc^2...”
how do we integrate csc^2 (x)*sec^2 (x)?
Button navigates to signup pageComment on 1151104828's post “how do we integrate csc^2...”
(0 votes)
Answer
- Mohit Jassi
  Posted 6 years ago. Direct link to Mohit Jassi's post “Let, I = ∫csc2x.sec2x.dx ...”
  Let,
  I = ∫csc2x.sec2x.dx
  = ∫ (1/ sin〗^2⁡x ).(1/〖cos〗^2⁡x ).dx
  = 4 ∫ 1/〖4 sin〗^2⁡〖x .〖cos〗^2⁡x 〗 dx
  = 4 ∫ 1/(2sinx.cosx)^2 .dx
  = 4 ∫ 1/(sin2x)^2 dx
  = 4 ∫ csc22x.dx
  = 4 ((-cot2x))/2 + C
  = - 2cot2x + C
  = - 2 ((1-〖tan〗^2⁡x)/2tanx) + C
  = - ((1-〖tan〗^2⁡〖x)〗)/tanx + C
  = (〖tan〗^2⁡x - 1)/tanx + C
  I = tanx – cotx + C
  Button navigates to signup page
  (3 votes)
vitocgiov
Posted 7 years ago. Direct link to vitocgiov's post “how do we integrate (x^2)...”
how do we integrate (x^2)/sqrt((x^2)-64)
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(0 votes)
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Sarvesh Yadav
Posted 6 years ago. Direct link to Sarvesh Yadav's post “(x)Ke Power -3/4(4x Squar...”
(x)Ke Power -3/4(4x Square +7x+xKe Power -1/4
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(0 votes)
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Amin Bou Hamdan
Posted 6 years ago. Direct link to Amin Bou Hamdan's post “Hello! How to derive or a...”
Hello! How to derive or anti-derive "log" and "ln"...
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- Mark Geary
  Posted 6 years ago. Direct link to Mark Geary's post “There are several videos ...”
  There are several videos and exercises on this topic. Use the "Search" function at the top of every page to look for the answer to your question.
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  (1 vote)
neenavazhavatta001
Posted 7 years ago. Direct link to neenavazhavatta001's post “let Gama(t)=2e^it for 0<=...”
let Gama(t)=2e^it for 0<=t<=2*3.14.then the value of the integral 1/2*3.14*I integral Gama (3z+1 ) dz /z(z-1) is
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