- -1 - tan x tanx - - dx-is equal to

We have tan (x + α x) = tan(x+ α) tan x/1 + tan x tan(x+ α) Then, we have ∫ [1 + tan x tan(x + α)]dx = ∫ cot α [tan(x + α) tan x]dx = cot α [ ln| cos (x ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 1

∫ [1 + tan x ...

Question

$\int [1 + t a n x t a n (x + α)] d x$ ,is equal to

t a n α I n | \frac{s i n (x + α)}{s i n x} | + C

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c o t α I n | \frac{s i n (x + α)}{s i n x} | + C

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c o t α I n | \frac{s i n x}{s i n (x + α)} | + C

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c o t α I n | \frac{c o s x}{c o s (x + α)} | + C

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Solution

The correct option is D $c o t α I n | \frac{c o s x}{c o s (x + α)} | + C$
We have $t a n (x + α - x) = \frac{t a n (x + α) - t a n x}{1 + t a n x t a n (x + α)}$
Then, we have
$\int [1 + t a n x t a n (x + α)] d x = \int c o t α [t a n (x + α) - t a n x] d x = c o t α [- l n | c o s (x + α) + l n | c o x x | |] + C = c o t α l n | \frac{c o s x}{c o s (x + α)} | + C$

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∫[1+tan xtan(x+α)]dx,is equal to

The correct option is D cot α In|cosxcos(x+α)|+CWe have tan(x+α−x)=tan(x+α)−tanx1+tanx tan(x+α) Then, we have ∫[1+tan x tan(x+α)]dx=∫cotα[tan(x+α)−tanx]dx=cot α[−ln|cos(x+α)+ln|cox x||]+C=cot α ln|cosxcos(x+α)|+C

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