Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you!
1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx
It says u = tan x to substitute
So if I use u = tan x, then my du = secx^2
then I have integral of (u^2)
then (1/3)u^3.....to get (1/3)tanx^3 then solve as 0 to pi/4 .......is that right?
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2. integral from 0 to 2root3 of (x^3)/(root(16-x^2))
I am pretty sure this has something to do with trig substitution. is it something with tan-1 or cos(theta)?
I am not sure how to go about this one at all.
Thanks again! =)
. integral from 0 to pi/4 of (tanx^2)(secx^4)dx
It says u = tan x to substitute
So if I use u = tan x, then my du = secx^2
no. du= sec^2x dx
then I have integral of (u^2)
Hmmm. INT u^2 sec^4x du/sec^2x
INT u^2 sec^2x du
how are you going to get rid of the sec^2 x?
second: draw the imaginary triangle, 4 is the hypotenuse, with angle theta , x is the opposite side, the adjacent side is sqrt (16-x^2)
then (1/3)u^3.....to get (1/3)tanx^3 then solve as 0 to pi/4 .......is that right?
For the first problem, you correctly identified that substituting u = tan(x) would be a good choice. However, the differential of u, du, is not equal to sec(x)^2. The correct differential is du = sec(x)^2 dx.
When substituting u = tan(x), you can rewrite the integral as ∫(tan(x)^2)(sec(x)^4) dx = ∫u^2 sec(x)^2 du.
To simplify this further, recall the trigonometric identity: tan^2(x) + 1 = sec^2(x). Rearranging this identity, we have sec^2(x) - 1 = tan^2(x).
Replacing sec^2(x) in the original integral with tan^2(x) + 1, we get ∫u^2 (tan^2(x) + 1) du.
Expanding this, we have ∫(u^2 tan^2(x) + u^2) du = ∫u^2 tan^2(x) du + ∫u^2 du.
The integral of u^2 du is straightforward and equals (1/3)u^3.
For the integral of u^2 tan^2(x) du, you can use the fact that tan(x) = u. Therefore, tan^2(x) = u^2.
Substituting these values back in, we have ∫u^2 tan^2(x) du = ∫u^2 u^2 du = ∫u^4 du = (1/5)u^5.
Adding these two results together, the final integral is (1/3)u^3 + (1/5)u^5.
You can now evaluate this integral from 0 to π/4 by subbing back in the value of u = tan(x).
For the second problem, it does involve trigonometric substitution. Let's rewrite the integral as ∫(x^3)/(√(16-x^2)) dx.
To find the appropriate substitution, observe that if you let x = 4sin(θ), then √(16-x^2) becomes √(16 - (4sin(θ))^2) = √(16 - 16sin^2(θ)) = √(16cos^2(θ)) = 4cos(θ).
Taking the derivative of x = 4sin(θ) with respect to θ, we get dx = 4cos(θ) dθ.
Now substitute these values back into the integral: ∫((4sin(θ))^3)/(4cos(θ)) * 4cos(θ) dθ.
Simplifying further, we have ∫ 16sin^3(θ) dθ.
As for the bounds of integration, x ranges from 0 to 2√3. To find the corresponding values of θ, you can use the substitution x = 4sin(θ).
When x = 0, θ = 0, and when x = 2√3, θ = π/3.
So the new integral becomes ∫[0, π/3] 16sin^3(θ) dθ.
Now, using a trigonometric identity, sin^3(θ) = (3sin(θ) - sin(3θ))/4, we can rewrite the integral as ∫[0, π/3] 16(3sin(θ) - sin(3θ))/4 dθ.
Simplifying further, we have ∫[0, π/3] 12sin(θ) - 4sin(3θ) dθ.
After integrating term by term, the final result is [12(-cos(θ)) + (cos(3θ))/3] evaluated from 0 to π/3.
By plugging in the bounds, you can find the exact value of the integral.