-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 ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 (sin x - x cos θ) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (sin x + 2 x cos θ) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (sin x - 2 x cos θ) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

\int \frac{cos x + \sqrt{3}}{1 + 4 sin (x + \frac{π}{3}) + 4 {sin}^{2} (x + \frac{π}{3})} d x

is
where

c

is constant of integration

\int \frac{(cos x + \sqrt{3}) d x}{1 + 4 sin (x + \frac{π}{3}) + 4 {sin}^{2} (x + \frac{π}{3})} =

\int \frac{sin \frac{5 x}{2}}{sin \frac{x}{2}} d x

is equal to:
(where

c

is a constant of integration.)

\int \frac{3 + 2 c o s x}{(2 + 3 c o s x)^{2}} d x

is equal to

Q. Solve

\frac{\sqrt{3}}{2} sin x - cos x = c o s^{2} x

Join BYJU'S Learning Program

∫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

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