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:


ANSWER

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


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
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