How do you evaluate the integral from 0 to $\frac{π}{4}$ $pi/4$ of $\frac{1 + {cos}^{2} x}{{cos}^{2} x} d x$ $(1 + cos^2 x) / (cos^2 x) dx$ ?

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How do you evaluate the integral from 0 to $\frac{π}{4}$ $pi/4$ of $\frac{1 + {cos}^{2} x}{{cos}^{2} x} d x$ $(1 + cos^2 x) / (cos^2 x) dx$ ?

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Answer 1 · 2015-04-10T11:29:29

Just do simple math : $\int_{0}^{\frac{π}{4}} \frac{1 + {cos}^{2} (x)}{{cos}^{2} (x)} d x$ $int_0^(pi/4)(1+cos^2(x))/cos^2(x) dx$

$= \int_{0}^{\frac{π}{4}} \frac{1}{{cos}^{2} (x)} + 1 d x$ $=int_0^(pi/4)1/cos^2(x)+1 dx$

$= (tan (x) + x) ⌉_{0}^{\frac{π}{4}}$ $=(tan(x)+x)~|_0^(pi/4)$

$= 1 + \frac{π}{4}$ $=1+pi/4$

EDIT : Here the "Funny" answer !

$\int_{0}^{\frac{π}{4}} \frac{1 + {cos}^{2} (x)}{{cos}^{2} (x)} d x = \int_{0}^{\frac{π}{4}} \frac{2 + {cos}^{2} (x) - 1}{{cos}^{2} (x)} d x$ $int_0^(pi/4)(1+cos^2(x))/cos^2(x)dx = int_0^(pi/4)(2+cos^2(x)-1)/cos^2(x)dx$

Factorize : $\int_{0}^{\frac{π}{4}} \frac{- (1 - {cos}^{2} (x)) + 2}{{cos}^{2} (x)} d x = \int_{0}^{\frac{π}{4}} \frac{- {sin}^{2} (x) + 2}{{cos}^{2} (x)} d x$ $int_0^(pi/4)(-(1-cos^2(x))+2)/cos^2(x)dx = int_0^(pi/4)(-sin^2(x)+2)/cos^2(x)dx$

$= - \int_{0}^{\frac{π}{4}} \frac{{sin}^{2} (x)}{{cos}^{2} (x)} d x + 2 \int_{0}^{\frac{π}{4}} \frac{1}{{cos}^{2} (x)} d x$ $=-int_0^(pi/4)sin^2(x)/cos^2(x) dx+2int_0^(pi/4)1/cos^2(x) dx$

$= - \int_{0}^{\frac{π}{4}} 1 + {tan}^{2} (x) - 1 d x + 2 \int_{0}^{\frac{π}{4}} \frac{1}{{cos}^{2} (x)} d x$ $=-int_0^(pi/4)1+tan^2(x)-1 dx + 2int_0^(pi/4)1/cos^2(x)dx$

$= - (tan (x) - x) ⌉_{0}^{\frac{π}{4}} + 2 (tan (x)) ⌉_{0}^{\frac{π}{4}}$ $=-(tan(x)-x)~|_0^(pi/4)+2(tan(x))~|_0^(pi/4)$

$= - 1 + \frac{π}{4} + 2 = 1 + \frac{π}{4}$ $=-1+pi/4+2 = 1+pi/4$

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How do you evaluate the integral from 0 to $\frac{π}{4}$ $pi/4$ of $\frac{1 + {cos}^{2} x}{{cos}^{2} x} d x$ $(1 + cos^2 x) / (cos^2 x) dx$ ?

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