Consider a function f(x) which satisfies f'(x)=tan2x+∫−π4π4f(x)dx. Also f(π4)=−π4. If the value of 8+π2π∫−π4π4f(x)dx=m, then find the value of m2

f'(x)=tan2x+K where K=∫−π4π4f(x)dx f(x)=tanx−x+Kx+C f(π4)=1−π4+Kπ4+C=−π4 C+1=−Kπ4 f(x)=tanx−x+Kx−Kπ4−1 Now, K=∫−π4π4(tanx−x+Kx⏟odd function−Kπ4−1)dx=−π2−Kπ28 hence, K=−4π8+π2 ∴8+π2π.∫−π4π4f(x)dx=−4≡m⇒m2=16

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Consider a function $f (x)$ which satisfies $f' (x) = \tan^{2} x + \int_{\frac{- π}{4}}^{\frac{π}{4}} f (x) d x .$ Also $f (\frac{π}{4}) = \frac{- π}{4}$ . If the value of $\frac{8 + π^{2}}{π} \int_{\frac{- π}{4}}^{\frac{π}{4}} f (x) d x = m,$ then find the value of $m^{2}$

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answer is 16.

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Detailed Solution

$f' (x) = \tan^{2} x + K w h e r e K = \int_{\frac{- π}{4}}^{\frac{π}{4}} f (x) d x f (x) = \tan x - x + K x + C f (\frac{π}{4}) = 1 - \frac{π}{4} + \frac{K π}{4} + C = \frac{- π}{4} C + 1 = \frac{- K π}{4} f (x) = \tan x - x + K x - \frac{K π}{4} - 1 N o w, K = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\underset{o d d f u n c t i o n}{\underset{⏟}{\tan x - x + K x}} - \frac{K π}{4} - 1) d x = \frac{- π}{2} - \frac{K π^{2}}{8} h e n c e, K = \frac{- 4 π}{8 + π^{2}} ∴ \frac{8 + π^{2}}{π} . \int_{\frac{- π}{4}}^{\frac{π}{4}} f (x) d x = - 4 \equiv m \Rightarrow m^{2} = 16$