Mathematics

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


ANSWER

$$\dfrac{\pi}{4}$$


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

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

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

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

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

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

 

Hence, this is the answer.

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