Question

$\dag \underline{\red{\mathfrak{Integration}}} :$

$\bf Evaluate : \displaystyle \int\left(\dfrac{x^2+\sin^2 x}{1+x^w\ \textless \ br /\ \textgreater \ 2}\right)\sec^2 x dx.$

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Answer 1

$\sf \large\underline{Correct\: Question} :$
$\bf \longmapsto Evaluate :\displaystyle\int\left(\dfrac{x^2+\sin^2 x}{1+x^2}\right)\sec^2 x dx$

$\sf\large\underline{Solution}:$

$\implies\displaystyle\int\left(\dfrac{x^2+\sin^2 x}{1+x^2}\right)\sec^2 x dx\\\\\implies \displaystyle\int\left(\dfrac{x^2+1 - \cos^2 x}{1+x^2}\right)\sec^2 x dx \\\\\blue{(\because \sin^2x=1-\cos^2x)} \\\\\implies \displaystyle\int\left(\dfrac{x^2+1}{1+x^2}-\dfrac{\cos^2x}{1+x^2}\right)\sec^2x dx\\\\\implies \displaystyle\int\left(\cancel{\dfrac{1+x^2}{1+x^2}} \sec^2x-\dfrac{\cos^2x \sec^2x}{1+x^2}\right)dx\\\\\blue{(\cos\theta\times\sec\theta =1)} \\\\\implies\displaystyle\int\sec^2x dx -\int\dfrac{1}{1+x^2}dx \\\\\implies \sf\tan x - tan^{-1}x+ C$

$\large\tt\underline{Therefore},\\\\ \longmapsto\displaystyle\int\left(\dfrac{x^2+\sin^2 x}{1+x^2}\right)\sec^2 x dx = \sf\tan x - tan^{-1}x+ C$

$\sf\pink{Some\: Standard\: Integrals:}$
⠀ ⠀• ∫ Sin x dx = -Cos x + C.
⠀ ⠀• ∫ Cos x dx = Sin x + C.
⠀ ⠀• ∫ Sec² x dx = tan x + C.
⠀ ⠀• ∫ cosec²x dx = -cot x + C.
⠀ ⠀• ∫ sec x tan x dx = sec x + C.
⠀ ⠀• ∫ cosec x cot x dx = - cosec x + C.
$\sf\quad\:\: \small{\bullet}\int x^n dx = \dfrac{x^{(n+1)}}{n+1} + C , n\neq -1 ( Power\:rule)$

Answer 2

Answer:

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