If fx - - ex 2-1 - tan x - tan2 x - -4 dx- where f3 -4 - 0- Then the value of lnf- is

F(x) = ∫ e^x (1 + tan x/1 tan x + ^2 (x + π/4)) dx = ∫ e^x (tan (π/4 + x) + ^2 (x + π/4)) dx ⇒ f(x) = e^x tan (π/4 + x) + C f(3 π/4) = 0 ⇒ C = 0 f (π) = e^π⇒ ...

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Byju's Answer

Standard XII

Mathematics

Integration of Trigonometric Functions

If fx = ∫ ex ...

Question

If $f (x) = \int e^{x} (\frac{2}{1 - tan x} + {tan}^{2} (x + \frac{π}{4})) d x$ , where $f (\frac{3 π}{4}) = 0.$ Then the value of $ln (f (π))$ is

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Solution

$f (x) = \int e^{x} (\frac{1 + tan x}{1 - tan x} + {sec}^{2} (x + \frac{π}{4})) d x$
$= \int e^{x} (tan (\frac{π}{4} + x) + {sec}^{2} (x + \frac{π}{4})) d x$
$\Rightarrow f (x) = e^{x} tan (\frac{π}{4} + x) + C$
$f (\frac{3 π}{4}) = 0 \Rightarrow C = 0$
$f (π) = e^{π} \Rightarrow ln (f (x)) = π$

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If f(x)=∫ex(21−tanx+tan2(x+π4)) dx, where f(3π4)=0. Then the value of ln(f(π)) is

f(x)=∫ex(1+tan x1−tanx+sec2(x+π4)) dx =∫ex(tan(π4+x)+sec2(x+π4)) dx ⇒f(x)=ex tan(π4+x)+C f(3π4)=0⇒C=0 f(π)=eπ⇒ln(f(x))=π

If $f (x) = \int e^{x} (\frac{2}{1 - tan x} + {tan}^{2} (x + \frac{π}{4})) d x$ , where $f (\frac{3 π}{4}) = 0.$ Then the value of $ln (f (π))$ is