$\int\limits_{0}^{\frac{\pi }{6}}tan^2(x)sec^2(x)\cdot dx
\\------------------\\
u=tan(x)\implies \frac{du}{dx}=sec^2(x)\implies \frac{du}{sec^2(x)}=dx
\\------------------\\
\int\limits_{0}^{\frac{\pi }{6}}u^2sec^2(x)\cdot \cfrac{du}{sec^2(x)}\implies 
\int\limits_{0}^{\frac{\pi }{6}}u^2\cdot du
\\ \quad \\
$

$\textit{now, we need to change the bounds as well, so}
\\------------------\\
u(0)=tan(0)\implies 0
\\ \quad \\

u\left( \frac{\pi }{6} \right)=tan\left( \frac{\pi }{6} \right)\implies \frac{1}{\sqrt{3}}
\\------------------\\
thus\implies \int\limits_{0}^{\frac{1 }{\sqrt{3}}}u^2\cdot du$

and surely you can take it from there,
recall, that, since we changed the bounds, with the u(x),
you don't need to change the variable "u", and simply,
get the integral of it, simple enough, and apply those bounds

5 0

3 years ago

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Find the value of x in this image

dangina [55]

2x+2x+(x-15)=180
4x+x-15=180
5x=180+15
5x=195
x=195/5
x=39

Good luck

4 0

3 years ago

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