Question

Trig substitution for int(9/(sqrt(3+x^2)))dx
the answer is 9ln(x+sqrt(3+x^2))+C.

Answer 1

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{9}{\sqrt{3+x^2}}~dx}\)

Answer 2

like this?

Answer 3

yeah

Answer 4

\(\large\color{black}{\displaystyle x=\sqrt{3}\cdot \tan \theta}\)

Answer 5

yeah I got that

Answer 6

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{9}{\sqrt{3+\left(\sqrt{3}\cdot \tan \theta\right)^2}}~\left(\sqrt{3}\cdot \sec^2\theta\right)d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{9\sqrt{3}\cdot \sec^2\theta}{\sqrt{3+3\tan ^2\theta}}~d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{9\sqrt{3}\cdot \sec^2\theta}{\sqrt{3}\sqrt{1+\tan ^2\theta}}~d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{9\sec^2\theta}{\sqrt{\sec ^2\theta}}~d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{9\sec^2\theta}{\sec \theta}~d\theta}\)

\(\large\color{black}{\displaystyle9\int\limits_{~}^{~}\sec\theta~d\theta}\)

Answer 7

I am not supposed to solve it for you, so try to work out the solution please.
I will keep typing and I will store it somewhere

Answer 8

You can't tell me the integral for sec \(\theta\) ?

Answer 9

ln (tan x + sec x)

Answer 10

yes, correct

Answer 11

yeah I got it from here

Answer 12

but in this case, NOT X, but THETA

Answer 13

and +C

Answer 14

yeah theta

Answer 15

Now, you have to solve for theta in terms of x, and substitute the original variable.

Answer 16

\(\large\color{black}{\displaystyle9\int\limits_{~}^{~}\sec\theta~d\theta= \\[0.9em] 9\ln\left(\tan \theta+\sec \theta\right)+C=\\[0.9em]
9\ln\left(\tan \left(?\right)+\sec \left(?\right)\right)+C}\)

\(\large\color{blue}{\displaystyle x=\sqrt{3}\cdot \tan \theta}\)
\(\large\color{blue}{\displaystyle x/\sqrt{3}=\tan \theta}\)
\(\large\color{blue}{\displaystyle \tan^{-1}\left(x/\sqrt{3}\right)= \theta}\)

\(\large\color{red}{\displaystyle9\ln\left(\tan \left(\tan^{-1}\left(x/\sqrt{3}\right)\right)+\sec \left(\tan^{-1}\left(x/\sqrt{3}\right)\right)\right)+C}\)


\(\large\color{black}{\displaystyle9\ln\left(\frac{x}{\sqrt{3}}+\sec \left(\tan^{-1}\left(x/\sqrt{3}\right)\right)\right)+C}\)


\(\large\color{black}{\displaystyle9\ln\left(\frac{x}{\sqrt{3}}+\sqrt{1+\arctan^2 \left(\tan^{-1}\left(x/\sqrt{3}\right)\right)}\right)+C}\)


\(\large\color{black}{\displaystyle9\ln\left(\frac{x}{\sqrt{3}}+\sqrt{1+\left(\frac{x}{\sqrt{3}}\right)^2}\right)+C}\)


\(\large\color{black}{\displaystyle9\ln\left(\frac{x}{\sqrt{3}}+\sqrt{1+\frac{x^2}{3}}\right)+C}\)


\(\large\color{black}{\displaystyle9\ln\left(\frac{x}{\sqrt{3}}+\sqrt{\frac{x^2+3}{3}}\right)+C}\)


\(\large\color{black}{\displaystyle9\ln\left(\frac{x}{\sqrt{3}}+\frac{\sqrt{x^2+3}}{\sqrt{3}}\right)+C}\)



\(\large\color{black}{\displaystyle9\ln\left(\frac{x+\sqrt{x^2+3}}{\sqrt{3}}\right)+C}\)


\(\large\color{black}{\displaystyle9\ln\left(x+\sqrt{x^2+3}\right)-9\ln\left(\sqrt{3}\right)+C}\)

9ln(√3) is also a constant, and an arbitrary constant minus 9ln(√3),
will still yield an arbitrary constant. THereofre you have:

\(\large\color{black}{\displaystyle9\ln\left(x+\sqrt{x^2+3}\right)+C}\)

Answer 17

Yeah I was making sec theta into 1/cos theta. Then integrating thanks

Answer 18

yes, that is certainly the way to show the integral of secθ,
knowing the fact that:

sec θ = cos θ / (1 - sin\(^2\)θ)

Answer 19

then u=sin θ
and so on...