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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.

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 class=

Respuesta :

[tex]\bf \displaystyle \int~\cfrac{sec^2(x)}{\sqrt{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)}}{\sqrt{u}}\cdot\cfrac{du}{\underline{sec^2(x)}}\implies \int~\cfrac{1}{\sqrt{u}}\cdot du\implies \int~u^{-\frac{1}{2}}\cdot du \\\\\\ 2u^{\frac{1}{2}}\implies 2\sqrt{1+tan(x)}+C[/tex]
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Answer:

[tex]2\sqrt{1+tan(x)} +C[/tex]

Step-by-step 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.