If \( \int\displaystyle\frac{\sin^4x}{\cos^8x}\,dx=a\tan^7x+b\tan^5x+C \), where \'C\' is an arbitrary constant then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5b=7a \)
Writing the integrand in terms of tan x and sec x brings out \( \sec^2x \), the derivative of tan x:
\( \displaystyle \frac{\sin^4x}{\cos^8x} \)
\( \displaystyle =\frac{\sin^4x}{\cos^4x}\cdot\frac{1}{\cos^4x} \)
\( \displaystyle =\tan^4x\sec^4x \)
\( \displaystyle =\tan^4x\,(1+\tan^2x)\sec^2x \)
Let t = tan x, so dt = sec2x dx:
\( \displaystyle \int(t^4+t^6)\,dt \)
\( \displaystyle =\frac{t^5}{5}+\frac{t^7}{7}+C \)
\( \displaystyle =\frac{1}{7}\tan^7x+\frac{1}{5}\tan^5x+C \)
\( \displaystyle a=\frac{1}{7},\quad b=\frac{1}{5} \)
\( \displaystyle 7a=1 \)
\( \displaystyle 5b=1 \)
So 5b = 7a.