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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 $f$ $\phantom{\rule{0.278em}{0ex}}⟺\phantom{\rule{0.278em}{0ex}}$ $f$ 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 $\int 2x\phantom{\rule{0.167em}{0ex}}dx$.
In general, $\int {f}^{\prime }\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=f\left(x\right)+c$.

## 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]
• 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).
• How should we integrate sin sqrt(x) using reverse chain rule?
• 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
• 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