The value of the integral ∫sinθ . sin2θsin6θ+sin4θ+sin2θ2sin4θ+3sin2θ+61−cos2θdθ
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
1189−2cos6θ−3cos4θ−6cos2θ32+c
b
11811−18cos2θ+9cos4θ−2cos6θ32+c
c
1189−2sin6θ−3sin4θ−6sin2θ32+c
d
11811−18sin2θ+9sin4θ−2sin6θ32+c
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
The given integral is ∫sinθ . sin2θsin6θ+sin4θ+sin2θ2sin4θ+3sin2θ+61−cos2θdθ=∫2sin2θsin6θ+sin4θ+sin2θ2sin4θ+3sin2θ+62sin2θ⋅cosθdθsubstitute sinθ=t,cosθ=dtThe given integral becomes=∫t6+t4+t22t4+3t2+6 dt=∫t5+t3+t2t6+3t4+6t2dtsubstitute 2t6+3t4+6t2=y⇒12t5+t3+tdt=dyhence, 112∫ydy=118y32+c=1182t6+3t4+6t232+c=1182sin6θ+3sin4θ+6sin2θ32+C=118−2cos6θ+9cos4θ−18cos2θ+1132+C