If \( \int\displaystyle\frac{\sin^6x}{\cos^{10}x}\,dx=-2a\tan^9x+b\tan^7x+c \) , ( where \( c \) is an arbitrary constant) , then which one of the following is TRUE?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 18a+7b=0 \)
The integrand can be written in tan and sec, keeping one \( \sec^2x \) for the substitution:
\( \displaystyle \frac{\sin^6x}{\cos^{10}x}=\frac{\sin^6x}{\cos^6x}\cdot\frac{1}{\cos^4x} \)
\( \displaystyle =\tan^6x\sec^4x \)
\( \displaystyle =\tan^6x(1+\tan^2x)\sec^2x \)
Let \( \tan x=t \), so \( \sec^2x\,dx=dt \):
\( \displaystyle \int(t^6+t^8)\,dt=\frac{t^7}{7}+\frac{t^9}{9}+c \)
\( \displaystyle =\frac{1}{9}\tan^9x+\frac{1}{7}\tan^7x+c \)
Comparing with \( -2a\tan^9x+b\tan^7x+c \):
\( \displaystyle -2a=\frac{1}{9}\ \Rightarrow\ a=-\frac{1}{18} \)
\( \displaystyle b=\frac{1}{7} \)
\( \displaystyle 18a=-1 \)
\( \displaystyle 7b=1 \)
So \( 18a+7b=0 \).