Find the integrals of the functions-cos 2 x-cos 2 -cos x-cos- d x

∫cos 2x cos 2 α/cos x cos αdx =∫(2 cos^2 x 1) (2 cos^2 α 1)/(cos x cos α)dx (∴ cos 2x=2 cos^2 x 1) =∫2 cos^2 x 1 2cos^2 α+1/(cos x cos α)dx =∫2 (cos^2 x cos^ ...

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Byju's Answer

Standard XII

Mathematics

Conditional Identities

Find the inte...

Question

Find the integrals of the functions.
$\int \frac{c o s 2 x - c o s 2 α}{c o s x - c o s α} d x .$

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Solution

$\int \frac{c o s 2 x - c o s 2 α}{c o s x - c o s α} d x = \int \frac{(2 c o s^{2} x - 1) - (2 c o s^{2} α - 1)}{(c o s x - c o s α)} d x (∴ c o s 2 x = 2 c o s^{2} x - 1)$

$= \int \frac{2 c o s^{2} x - 1 - 2 c o s^{2} α + 1}{(c o s x - c o s α)} d x = \int \frac{2 (c o s^{2} x - c o s^{2} α)}{(c o s x - c o s α)} d x$
$= 2 \int \frac{(c o s x - c o s α) (c o s x + c o s α)}{(c o s x - c o s α)} d x | ∴ a^{2} - b^{2} = (a + b) (a - b) |$
$= 2 [\int c o s x d x + c o s α \int 1 d x] = 2 [s i n x + c o s α . x] + C = 2 s i n x + 2 x c o s α + C$

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Find the integrals of the functions. ∫cos2x−cos2αcosx−cosαdx.

Find the integrals of the functions.
$\int \frac{c o s 2 x - c o s 2 α}{c o s x - c o s α} d x .$