Here's my first question: (FIrst part courtesy of Count Iblis)
Integral of x sqrt(19x-7)dx ?
Write the integral in terms of functins you do know the inegral of. Rewrite the factor of x as follows:
x = 1/19 (19 x) =
1/19 (19 x - 7 + 7) =
1/19 (19 x - 7) + 7/19
^ He wrote this down, but I don't know what to do with it. V_V
My next, and hopefully last question, is thus.
Integral of [(sin(7x)^2)*(sec(7x)^4) dt]
I flipped through my text book, and I can't find anything like it. According to my prof. I'm supposed to use Integration Tables.
<<Integral of x sqrt(19x-7)dx >>
Let u = 19x -7
dx = (1/19) du
x = (1/19)(u + 7)
So the integral becomes the sum of two integrals:
Integral of (1/19)u^(3/2) du
+ Integral of (7/19)u^(1/2) du
both of which are easy integrals. Remember to change from u back to (19x - 7) when you are done.
<<Integral of [(sin(7x)^2)*(sec(7x)^4) dt] >>
I assume your differential variable of integration is dx, not dt.
I suggest you rewrite sin(7x)^2 as 1 = cos^2(7x). That leaves you with two integrals: one involving sec^4(7x) and the other sec^2(7x) (since cos = 1/sec). Then substitute u for 7x and use a table of integrals. The integral of sec^2 du is tan u. There is a recursion formula for the integral of sec^n u du. It says that
Integral of sec^4 u du =
[sinu/(3 cos^3u)] + (2/3)tan u
Thank you so much. You cleared that up really well. Thank you.
To find the integral of x sqrt(19x-7) dx, we can use a technique called u-substitution. Here's how you can solve it:
Let u = 19x - 7. Then, differentiate both sides of the equation to find du.
du/dx = 19
Now, solve for dx:
dx = du/19
Substitute these expressions into the original integral:
∫ x sqrt(19x-7) dx = ∫ [(u + 7)/19] sqrt(u) (du/19)
Combine the constants:
(1/19^2) ∫ (u + 7) sqrt(u) du
Expand and distribute:
(1/19^2) ∫ (u^(3/2) + 7u^(1/2)) du
Integrate each term separately:
(1/19^2) ∫ u^(3/2) du + (1/19^2) ∫ 7u^(1/2) du
For the first term, u^(3/2), use the power rule for integration:
(1/19^2) * (2/5) * u^(5/2)
For the second term, 7u^(1/2), also use the power rule:
(1/19^2) * (2/3) * 7 * u^(3/2)
Simplify:
(2/95^2) * u^(5/2) + (14/3 * 19^2) * u^(3/2) + C
Replace u with 19x - 7:
(2/95^2) * (19x - 7)^(5/2) + (14/3 * 19^2) * (19x - 7)^(3/2) + C
This is the final answer to the integral of x sqrt(19x-7) dx.
As for your second question, the integral of [(sin(7x)^2)*(sec(7x)^4) dt], you mentioned that you couldn't find it in your textbook and your professor suggested using integration tables. While integration tables can be useful for certain functions, in this case, you don't necessarily need them.
You can simplify the expression using trigonometric identities. Start by using the identity sec^2(x) = 1 + tan^2(x), and rewrite sec^4(x) as (sec^2(x))^2. Then, substitute sec^2(x) = 1 + tan^2(x) into the integral:
∫ [(sin(7x)^2)*(sec(7x)^4)] dx = ∫ [(sin(7x)^2)*((1 + tan^2(7x))^2)] dx
Expand the square of (1 + tan^2(7x)):
∫ [(sin(7x)^2)*(1 + 2tan^2(7x) + tan^4(7x))] dx
Now, distribute sin(7x)^2:
∫ [sin(7x)^2 + 2sin(7x)^2*tan^2(7x) + sin(7x)^2*tan^4(7x)] dx
Split the integral into three separate integrals:
∫ sin(7x)^2 dx + 2∫ sin(7x)^2*tan^2(7x) dx + ∫ sin(7x)^2*tan^4(7x) dx
The first integral, ∫ sin(7x)^2 dx, can be solved using the double-angle identity sin^2(x) = (1 - cos(2x))/2:
∫ [1 - cos(14x)]/2 dx
= (1/2) ∫ (1 - cos(14x)) dx
= (1/2) [x - (1/14) * sin(14x)] + C
For the second and third integrals, you can use a substitution, letting u = tan(7x):
For 2∫ sin(7x)^2*tan^2(7x) dx, substitute sin^2(7x) = 1 - cos^2(7x):
= 2∫ (1 - cos^2(7x)) * tan^2(7x) dx
= 2∫ (1 - u^2) du
= 2[u - (1/3)u^3] + C
And for ∫ sin(7x)^2*tan^4(7x) dx, use the same substitution but with an additional u^2 factor:
= ∫ (1 - cos^2(7x)) * tan^4(7x) dx
= ∫ (1 - u^2) * u^2 du
= [u^3/3 - u^5/5] + C
Finally, substitute back u = tan(7x) to get the final result for each integral and combine all the terms.
Note: The integrals involving trigonometric functions can be quite challenging and may require additional trigonometric identities or techniques such as integration by parts.