Question

Evaluate: $\int\sqrt[3]{\frac{sin^{2} \,x}{cos^{14}\, x}}dx$

• A. $-\frac{3}{5}tan ^{\frac{5}{3}} x -\frac{3}{11} tan ^{-\frac{11}{3}} +C$
• B. $\frac{3}{5}tan ^{\frac{5}{3}} x +\frac{3}{11} tan ^{\frac{11}{3}}x +C$
• C. $\frac{3}{5}tan ^{\frac{5}{3}} x -\frac{3}{11} tan ^{\frac{11}{3}}x +C$
• D. $\frac{3}{5}tan^{\frac{3}{5}}x +\frac{3}{11} tan ^{\frac{11}{3}}x +C$

Answer 1

Answer: (B)

Let $ I = \int\sqrt[3]{\frac{sin^{2}\, x}{cos^{14} \,x}}dx = \int sin^{\frac{2}{3}} \,x cos^{-\frac{14}{3}}x \,dx$

Here, the sum of the exponents of $sin\, x$ and $cos\, x$ is $-4$, which is a negative even integer. So, we divide and multiply by $cos^4 \,x$ to get

$I = \int sin^{\frac{2}{3}} \,x cos^{-\frac{14}{3}}x cos^{4}x sec^{4} x\, dx $

$\Rightarrow I = \int\frac{sin^{\frac{2}{3}} x}{cos^{\frac{2}{3}} x} sec^{4} x dx$

$\Rightarrow I = \int tan ^{\frac{2}{3}} x \left(1+tan^{2}x\right) sec^{2}x dx$

Put $tan x = t \Rightarrow sec^{2}x dx =dt $

$ I= \int\left(t^{\frac{2}{3}} +t^{\frac{8}{3}}\right)dt = \frac{3}{5}t^{\frac{5}{3}}+\frac{3}{11}t^{\frac{11}{3}}+C$

$ = \frac{3}{5} tan ^{\frac{5}{3}}x +\frac{3}{11} tan ^{\frac{11}{3}}x +C$