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bogdanovich [222]
3 years ago
6

Calculus: Help ASAP

Mathematics
2 answers:
levacccp [35]3 years ago
7 0

Answer:

2\sqrt{1+tan(x)} +C

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.

wariber [46]3 years ago
6 0
\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
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