Evaluate the integral of the quotient of the secant squared of x and the square root of the quantity 1 plus tangent x, dx. https://gyazo.com/56a83617dc456d74002487417af5f68a
Nice. I just started reviewing by calculus yesterday, after my last class was in 1977, yes the year. I am still practicing Calc I. So sorry, I'm to rusty for something this advance.
Refer to the attachment from Mathematica v9, home edition:
what is up with this attachement? If you don't know how to help the user don't just post the solution with answer from mathematica like this, because that's inapropriate.
Of course the steps are right and the quickest (mathematica, dah), but if you still need help or have questions, then reply.
Please explain how it was done.
Recall that \(\dfrac{\mathrm{d}}{\mathrm{d}x}\tan x=\sec^2x\), so a substitution of \(u=\tan x\) would make things simpler: \[\int\frac{\sec^2x}{\sqrt{1+\tan x}}\,\mathrm{d}x\stackrel{u\tan x}{=}\int \frac{\mathrm{d}u}{\sqrt{1+u}}\]
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