If ∫sin2⁡α−sin2⁡xcos⁡x−cos⁡αdx=f(x)+Ax+B and B ∈ R then

=sin2⁡x=1−cos2⁡x=∫1−cos2⁡α−1−cos2⁡xcos⁡x−cos⁡αdx=∫cos2⁡x−cos⁡ 2αcos⁡x−cos⁡αdx=∫(cos⁡x+cos⁡α)⋅dx=sin⁡x+(cos⁡α)⋅x+CSo, f(x)=sin⁡x and A=cos⁡α

If ∫sin2⁡α−sin2⁡xcos⁡x−cos⁡αdx=f(x)+Ax+B and B ∈ R then

=sin2⁡x=1−cos2⁡x=∫1−cos2⁡α−1−cos2⁡xcos⁡x−cos⁡αdx=∫cos2⁡x−cos⁡ 2αcos⁡x−cos⁡αdx=∫(cos⁡x+cos⁡α)⋅dx=sin⁡x+(cos⁡α)⋅x+CSo, f(x)=sin⁡x and A=cos⁡α

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\int \frac{\sin^{2} α - \sin^{2} x}{\cos x - \cos α} dx = f (x) + Ax + B$ and B $\in$ R then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

A = cos α

f(x) = sin x

A = 2cos α

f(x) = 2 sin x

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} = \sin^{2} x = 1 - \cos^{2} x \\ = \int \frac{1 - \cos^{2} α - (1 - \cos^{2} x)}{\cos x - \cos α} dx \\ = \int \frac{\cos^{2} x - \cos^{2} α}{\cos x - \cos α} dx \\ = \int (\cos x + \cos α) \cdot dx \\ = \sin x + (\cos α) \cdot x + C \\ So, f (x) = \sin x and A = \cos α \end{array}$