Mathematics

# $\displaystyle\int {\frac{{\left( {{x^2} + {{\sin }^2}x} \right){{\sec }^2}x}}{{\left( {1 + {x^2}} \right)}}dx}$

##### SOLUTION
$I = \int {\frac{{\left( {{x^2} + {{\sin }^2}x} \right){{\sec }^2}x}}{{\left( {1 + {x^2}} \right)}}dx}$
$= \int {\left[ {\frac{{\left( {{{\sec }^2}x{{\sin }^2}x + {{\sec }^2}x} \right)}}{{\left( {1 + {x^2}} \right)}} + {{\sec }^2}x} \right]dx}$
$= \int {\left( {{{\sec }^{2}x} - \frac{1}{{{x^2} + 1}}} \right)} dx$
$= \tan x - {tan^4}x + C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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