Mathematics

If $f(x)=\int { \left( \dfrac { x^2+\sin^2x }{ 1+x^2 } \right) } \sec^2 x dx$ and f(0)=0,then  f (1) equals:

$\tan 1-\dfrac { \pi }{ 4 }$

SOLUTION
$f(x)=\int { \left( \dfrac { x^2+sin^2x }{ 1+x^2 } \right) } sec^2x dx$

$= \int { \left( \dfrac { x^2+1-(1-sin^2x) }{ 1+x^2 }sec^2x \right) } dx$

$=\int (sec^2x-\dfrac {cos^2x sec^2x} {1+x^2})dx$
$=\int sec^2x- \int \dfrac {1} {1+x^2} dx$

$f(x)= tanx - tan^{-1}x+ \alpha$

$f(0)=0\Rightarrow \alpha=0$

$f(1)=tan 1 -\dfrac { \pi }{ 4 }$

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

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