Differentiate ln [x+√(x^2+1)] with respect to x
hence evaluate limit 0 to 1
∫ 1/√(x^2+1) dx
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Use the identity cos3x=4cos^3x-3cosx,
find ∫cos^3x dx
* Is there a printing error in the question? the limit is from pi/2 to pi/2?!
To differentiate ln [x+√(x^2+1)] with respect to x, we can use the chain rule. Let's break it down step by step:
Step 1: Identify the function and the inner function.
The function we need to differentiate is ln [x+√(x^2+1)]. The inner function is x+√(x^2+1).
Step 2: Differentiate the inner function.
To differentiate x+√(x^2+1), we treat the square root as a power of 1/2. The derivative of x is 1, and the derivative of √(x^2+1) is (1/2)*(2x)/(√(x^2+1)). Simplifying this gives us (1+x)/(√(x^2+1)).
Step 3: Apply the chain rule.
Using the chain rule, we multiply the derivative of the outer function (ln) with the derivative of the inner function:
d/dx [ln [x+√(x^2+1)]] = (1+x)/(√(x^2+1)) * 1/(x+√(x^2+1))
This derivative is the final result of differentiating ln [x+√(x^2+1)] with respect to x.
Now, let's move on to evaluating the integral:
To evaluate the integral ∫ 1/√(x^2+1) dx from 0 to 1, we can use the substitution method. Let u = x^2 + 1, then du = 2x dx.
Substituting these values gives us:
∫ 1/√(x^2+1) dx = ∫ 1/√u * (1/2x) du
= (1/2) ∫ 1/u du
Integrating 1/u gives us ln|u|:
∫ 1/u du = ln|u| + C
Now substitute back in for u:
ln|x^2 + 1| + C
To determine the value of the definite integral from 0 to 1, we evaluate the antiderivative at the upper and lower limits:
∫[0,1] 1/√(x^2+1) dx = [ln|(1^2+1)| - ln|(0^2+1)|]
= ln|2| - ln|1|
= ln(2) - ln(1)
= ln(2)
Therefore, the value of the integral from 0 to 1 of 1/√(x^2+1) is ln(2).
Moving on to the next question about ∫cos^3x dx:
We will first use the provided identity: cos3x = 4cos^3x - 3cosx.
Rearranging the identity, we get:
cos^3x = (1/4)cos3x + (3/4)cosx
Now, we can rewrite the integral as:
∫cos^3x dx = ∫[(1/4)cos3x + (3/4)cosx] dx
Considering each term separately:
∫(1/4)cos3x dx = (1/4)∫cos3x dx
This is a straightforward integral of cos3x, which gives us:
(1/4) * (1/3)sin3x + C1 = (1/12)sin3x + C1
∫(3/4)cosx dx = (3/4)∫cosx dx
This is a simple integral of cosx, giving us:
(3/4)sinx + C2
Putting it all together, we have:
∫cos^3x dx = (1/12)sin3x + (3/4)sinx + C
Regarding the second part of the question about the limits from pi/2 to pi/2, it does seem like there is a printing error. The integral is evaluated over a range, and the limits should represent a lower and upper bound. In this case, since the limits are the same value, the integral has no region to evaluate and the result is 0.