Question

Calculus: Help ASAP

Evaluate the integral of the quotient of the secant squared of x and the square root of the quantity 1 plus tangent x, dx.

Answer 1

`2√(1+tan(x)) +C`

Explanation:

To start solving this you need to use substitution. I let u = 1+tan(x). Next you need to find du/dx, which is sec^2(x) using trigonometric properties. Solve for dx and get dx = du / sec^2(x). Next put the new dx back in. This gives you integral sec^2(x) / sqrt u * du / sec^2(x). The sec^2(x) cancels and the new expression is integral 1/sqrt u * du, which can be simplified to integral u^-1/2 * du. You then take the integral and get 2u^1/2. Lastly, substitute the original u back in and get 2 sqrt 1+tan(x) + C.

Answer 2

`\bf \displaystyle \int~\cfrac{sec^2(x)}{√(1+tan(x))}\cdot dx\\\\ -------------------------------\\\\ u=1+tan(x)\implies \cfrac{du}{dx}=sec^2(x)\implies \cfrac{du}{sec^2(x)}=dx\\\\ -------------------------------\\\\ \displaystyle \int~\cfrac{\underline{sec^2(x)}}{√(u)}\cdot\cfrac{du}{\underline{sec^2(x)}}\implies \int~\cfrac{1}{√(u)}\cdot du\implies \int~u^{-(1)/(2)}\cdot du \\\\\\ 2u^{(1)/(2)}\implies 2√(1+tan(x))+C`