-cos 2 x-cos 2 -cos x-cos- d x is equal toA- 2sin x-x cos-CB- 2sin x-x cos-CC- 2sin x-2 x cos-CD- 2sin x-2 x cos-C

The correct option is A 2(sin x+xcosθ)+C∫cos 2x cos 2θ/cos x cosθdx = ∫2cos^2 x 1 2cos^2θ+1/cos x cosθdx = ∫2(cos x+cosθ)(cos x cosθ)/cos x cosθdx = ∫2( cos ...

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Byju's Answer

Standard XII

Mathematics

Theorems on Integration

∫cos 2 x-cos ...

Question

$\int \frac{cos 2 x - cos 2 θ}{cos x - cos θ} d x$ is equal to

2 (sin x + x cos θ) + C

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2 (sin x - x cos θ) + C

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2 (sin x + 2 x cos θ) + C

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2 (sin x - 2 x cos θ) + C

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Solution

The correct option is A $2 (sin x + x cos θ) + C$
$\int \frac{cos 2 x - cos 2 θ}{cos x - cos θ} d x = \int \frac{2 {cos}^{2} x - 1 - 2 {cos}^{2} θ + 1}{cos x - cos θ} d x = \int \frac{2 (cos x + cos θ) (cos x - cos θ)}{cos x - cos θ} d x = \int 2 (cos x + cos θ) d x = 2 (sin x + x cos θ) + C$

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∫cos2x−cos2θcosx−cosθdx is equal to

The correct option is A 2(sinx+xcosθ)+C∫cos2x−cos2θcosx−cosθdx = ∫2cos2x−1−2cos2θ+1cosx−cosθdx= ∫2(cosx+cosθ)(cosx−cosθ)cosx−cosθdx= ∫2(cosx+cosθ)dx= 2(sinx+xcosθ)+C

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