Mathematics

# Solve $\displaystyle\int_0^{\pi /2} {} \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x}}dx$

$\dfrac{\pi}{4}$

##### SOLUTION

Consider the given integral.

$I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x+{{\cos }^{2}}x} \right)}dx$                …….. (1)

We know that

$\int_{b}^{a}{f\left( x \right)dx}=\int_{b}^{a}{f\left( a+b-x \right)dx}$

Therefore,

$I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x} \right)}dx$           …… (2)

On adding equation (1) and (2), we get

$2I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x} \right)}dx$

$2I=\int_{0}^{\dfrac{\pi }{2}}{1}dx$

$2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$2I=\dfrac{\pi }{2}-0$

$I=\dfrac{\pi }{4}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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$\displaystyle \int \tan x\log \sec xdx$
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• C. $\displaystyle \left ( \log \sec x \right )^{2}.$
• D. $\displaystyle \frac{1}{2}\left ( \log \sec x \right )^{2}.$

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