Find the value of I if I-0-4In1-tan x d x

I = ∫^π/4_0 In(1 + tan x)dx = ∫^π/4_0 In [1 + tan (π/4 x)dx [Using ∫^a_0 f(a x)dx] = ∫^π/4_0 In [1 + tan π/4 tan x/1 + tan π/4tan x]dx = ∫^π/4_0 In [1 + ...

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Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

Find the valu...

Question

Find the value of I if $I = \int_{0}^{\frac{π}{4}} I n (1 + t a n x) d x$

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Solution

$I = \int_{0}^{\frac{π}{4}} I n (1 + t a n x) d x = \int_{0}^{\frac{π}{4}} I n [1 + t a n (\frac{π}{4} - x) d x [U s i n g \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{4}} I n [1 + \frac{t a n \frac{π}{4} - t a n x}{1 + t a n \frac{π}{4} t a n x}] d x = \int_{0}^{\frac{π}{4}} I n [1 + \frac{1 - t a n x}{t a n x}] d x = \int_{0}^{\frac{π}{4}} I n [\frac{2}{1 + t a n x}] d x = \int_{0}^{\frac{π}{4}} [I n 2 - I n (1 + t a n x)] d x = \int_{0}^{\frac{π}{4}} I n 2 d x - \int_{0}^{\frac{π}{4}} I n (1 + t a n x) d x \Rightarrow I = (I n 2) [x]_{0}^{\frac{π}{4}} - I \Rightarrow 2 I = \frac{π}{4} I n 2 ∴ I = \frac{π}{8} I n 2$

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Find the value of I if I=∫π40In(1+tan x)dx

I=∫π40In(1+tan x)dx=∫π40In[1+tan(π4−x)dx[Using∫a0f(a−x)dx]=∫π40In[1+tanπ4−tanx1+tanπ4tan x]dx=∫π40In[1+1−tan xtan x]dx=∫π40In[21+tanx]dx=∫π40[In 2−In(1+tan x)]dx=∫π40In 2dx−∫π40In(1+tan x)dx⇒I=(In 2)[x]π40−I⇒2I=π4In 2∴I=π8In 2

Find the value of I if $I = \int_{0}^{\frac{π}{4}} I n (1 + t a n x) d x$