Value of \( \int\displaystyle \frac{dx}{\sin^{\frac{3}{5}}x\,.\,\cos^{\frac{7}{5}}x} \) is (where c is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{5}{2}\tan^{\frac{2}{5}}x+c \)
The powers add up to \( \frac{3}{5}+\frac{7}{5}=2 \), so dividing the numerator and denominator by \( \cos^2 x \) brings in tan x and \( \sec^2 x \):
\( \displaystyle \frac{1}{\sin^{3/5}x\,\cos^{7/5}x}=\frac{1}{\left(\frac{\sin x}{\cos x}\right)^{3/5}\cos^2 x} \)
\( \displaystyle =\frac{\sec^2 x}{\tan^{3/5}x} \)
Let t = tan x, so \( dt=\sec^2 x\,dx \):
\( \displaystyle \int t^{-3/5}\,dt=\frac{t^{2/5}}{2/5}+c \)
\( \displaystyle =\frac{5}{2}\tan^{2/5}x+c \)
Shortcut: Differentiate the expression in Option (1):
\( \displaystyle \frac{d}{dx}\left[\frac{5}{2}\tan^{2/5}x\right]=\frac{5}{2}\cdot\frac{2}{5}\tan^{-3/5}x\cdot\sec^2 x \)
\( \displaystyle =\frac{\sec^2 x}{\tan^{3/5}x} \)
\( \displaystyle =\frac{1}{\sin^{3/5}x\,\cos^{7/5}x} \)
This is the integrand, so Option (1) is correct.