I've got some work done on my own now and have started to resubstitute to get back to having the function in terms of x.
Can anyone tell me what tan * sec^-1(x) is equal to?
Now, once I perform that integral do I just resubstitute a lot to get to the correct answer? I mean, when finding the antiderivative of something which has a trig function in it, you almost always get something with a trig function still in it. I am still not seeing how this will lead to the...
My steps to get something with only cos(v) are as follows:
sin^2(v)/cos^2(v) * (1/cos(v))dv
(1-cos^2(v))/cos^2(v) * (1/cos(v))dv
(1-cos^2(v))/cos^3(v) dv
1/cos^3(v) - 1/cos(v)
Is this the right integral I should end up with?
I don't know what the general formula is... I just know that if you substitute in for u than your du value changes also. I have done substitution before and been fine, but I can't give you a general formula.
I'm completely lost still. I can't get past the answer I had before. I just don't understand how to do this combination of integration by parts AND trigonometric substitution.
The integral of tanudu is -ln|cos{u}|+C which then turns into -ln|cos{x+1}|+C.
This still isn't the right answer which can be seen in the previous quote by saladsamurai.
Something is still going wrong!
The Final Answer is:
\displaystyle{\frac{1}{2}}(x+1)\sqrt{x^2+2x}-\displaystyle{\frac{1}{2}}\ln{|x+1+\sqrt{x^2+2x}|}+C
This is in the form of
\int{udv}=uv-\int{vdu}
Which is valid for integration by parts.. I know we have to do trigonometric substitution also, but I don't know how we will...
So then with the U-Substitution,
We would have sec2u-1du beneath the square root. Does this then turn into tan2udu beneath the square root and then I perform the integral or try to do substitution again, etc.?
I have the answer (found in my textbook for the homework problem) which is:
1/2 * (x+1) * The Integral of: Square Root of (x^2 + 2x) - 1/2 the natural log of (x +1 + square root of (x^2 + 2x)) + C
Sorry... I have tried to do this even with the u substitution. I am just completely clueless...
Micromass -
When you say complete the square, what do you mean?
If I complete the square, that would give me the square root of (x+1)^2 - 1, right?
Then do I use trigonometic substitution?
Pardon my use of the program! I am new to Physics Forums!
Homework Statement
EVALUATE
The Integral of: Square root of (x^2 + 2x)
The Integral of: x * Square root of (x^2 + 7)
Homework Equations
Integrating by Parts Method
The Integral of udv = u*v - the integral of v*duThe Attempt at a...