Q.

If ∫sin32x+cos32xsin3xcos3xsin(x-θ) dx=Acosθtanx-sinθ+Bcosθ-sinθcotx+C, where C is the integration constant, then AB is equal to                       [2024]

1 4cosec(2θ)  
2 2secθ    
3 4secθ    
4 8cosec(2θ)  

Ans.

(4)

Let I=∫sin3/2x+cos3/2xsin3x·cos3x·sin(x-θ) dx

=∫dxcos3x·sin(x-θ)+∫dxsin3x·sin(x-θ)

=∫dxcos2xsinxcosθ-cosxsinθcosx+∫dxsin2xsinxcosθ-cosxsinθsinx

=∫sec2xtanxcosθ-sinθ dx+∫cosec2xcosθ-cotxsinθ dx

⇒I=I1+I2  (Let)                                    ...(i)

Now, I1=∫sec2x dxtanxcosθ-sinθ

Put tanx cosθ-sinθ=t⇒cosθ·sec2x dx=dt

∴  I1=∫dtcosθ·t⇒I1=2t1/2cosθ

⇒I1=2(tanx cosθ-sinθ)1/2cosθ

And, I2=∫cosec2x cosθ-cotx sinθdx

Put cosθ-cotxsinθ=v⇒sinθ cosec2x dx=dv

∴  I2=∫dvsinθv=2vsinθ⇒I2=2cosθ-cotxsinθsinθ

From (i),

I=2tanxcosθ-sinθcosθ+2cosθ-cotxsinθsinθ+C and

I=Acosθtanx-sinθ+Bcosθ-sinθcotx+C

So, A=2cosθ,  B=2sinθ

∴  AB=4·22sinθcosθ=8cosec2θ