Please can you help me with this question?
Evaluate this definite integral by using some suitable substitution
\(\displaystyle \L
\int_{{\textstyle{\pi \over 6}}}^{{\textstyle{\pi \over 3}}} {\tan ^2 x\sec ^2 x{\rm }dx}\)
\(\displaystyle \L
let{\rm }u = \tan x{\rm }du = \sec ^2 x{\rm }dx\)
\(\displaystyle \L
so,\int_{{\textstyle{\pi \over 6}}}^{{\textstyle{\pi \over 3}}} {u^2 du} = \frac{1}{3}u^3 = \frac{1}{3}\left[ {\tan ^3 x} \right]_{{\textstyle{\pi \over 6}}}^{{\textstyle{\pi \over 3}}}\)
\(\displaystyle \L
= \frac{1}{3}\left[ {3\sqrt 3 - \frac{{\sqrt 3 }}{9}} \right] = \frac{{26\sqrt 3 }}{{27}}\)
Can you tell me if this is the correct answer?
:?:
Evaluate this definite integral by using some suitable substitution
\(\displaystyle \L
\int_{{\textstyle{\pi \over 6}}}^{{\textstyle{\pi \over 3}}} {\tan ^2 x\sec ^2 x{\rm }dx}\)
\(\displaystyle \L
let{\rm }u = \tan x{\rm }du = \sec ^2 x{\rm }dx\)
\(\displaystyle \L
so,\int_{{\textstyle{\pi \over 6}}}^{{\textstyle{\pi \over 3}}} {u^2 du} = \frac{1}{3}u^3 = \frac{1}{3}\left[ {\tan ^3 x} \right]_{{\textstyle{\pi \over 6}}}^{{\textstyle{\pi \over 3}}}\)
\(\displaystyle \L
= \frac{1}{3}\left[ {3\sqrt 3 - \frac{{\sqrt 3 }}{9}} \right] = \frac{{26\sqrt 3 }}{{27}}\)
Can you tell me if this is the correct answer?
:?: