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## Class 12 math (India)

### Course: Class 12 math (India) > Unit 9

Lesson 3: Indefinite integrals of common functions- Indefinite integrals of sin(x), cos(x), and eˣ
- Indefinite integral of 1/x
- Indefinite integrals: eˣ & 1/x
- Particular solutions to differential equations: rational function
- Particular solutions to differential equations: exponential function
- Particular solutions to differential equations
- Indefinite integrals: sin & cos
- Integrating trig functions
- Antiderivatives and indefinite integrals review
- Common integrals review

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# Antiderivatives and indefinite integrals review

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 ${x}^{2}$ is $2x$ . This means that an antiderivative of $2x$ is ${x}^{2}$ .

${f}^{\prime}$ is the derivative of $f$ $\phantom{\rule{0.278em}{0ex}}}\u27fa{\textstyle \phantom{\rule{0.278em}{0ex}}$ $f$ is an antiderivative of ${f}^{\prime}$

Each function has a family of antiderivatives. For example, the antiderivatives of $2x$ 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 $2x$ is expressed as $\int 2x{\textstyle \phantom{\rule{0.167em}{0ex}}}dx$ .

In general, $\int {f}^{\prime}(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx=f(x)+c$ .

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

## Want to join the conversation?

- 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.(1 vote)
- 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](27 votes)

- 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?(1 vote)
- 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).(4 votes)

- How should we integrate sin sqrt(x) using reverse chain rule?(2 votes)
- 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(0 votes)

- how do we integrate csc^2 (x)*sec^2 (x)?(0 votes)
- Let,

I = ∫csc2x.sec2x.dx

= ∫ (1/ sin〗^2x ).(1/〖cos〗^2x ).dx

= 4 ∫ 1/〖4 sin〗^2〖x .〖cos〗^2x 〗 dx

= 4 ∫ 1/(2sinx.cosx)^2 .dx

= 4 ∫ 1/(sin2x)^2 dx

= 4 ∫ csc22x.dx

= 4 ((-cot2x))/2 + C

= - 2cot2x + C

= - 2 ((1-〖tan〗^2x)/2tanx) + C

= - ((1-〖tan〗^2〖x)〗)/tanx + C

= (〖tan〗^2x - 1)/tanx + C

I = tanx – cotx + C(3 votes)

- how do we integrate (x^2)/sqrt((x^2)-64)(0 votes)
- (x)Ke Power -3/4(4x Square +7x+xKe Power -1/4(0 votes)
- Hello! How to derive or anti-derive "log" and "ln"...(0 votes)
- 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.(1 vote)

- 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(0 votes)