Question

Evaluate: $\int \sqrt[3]{\frac{si{n}^{2}x}{co{s}^{14}x}}dx$

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

Answer 1

Answer: (B)

Let $I=\int \sqrt[3]{\frac{si{n}^{2}x}{co{s}^{14}x}}dx=\int si{n}^{\frac{2}{3}}xco{s}^{-\frac{14}{3}}xdx$

Here, the sum of the exponents of $sinx$ and $cosx$ is $-4$, which is a negative even integer. So, we divide and multiply by $co{s}^{4}x$ to get

$I=\int si{n}^{\frac{2}{3}}xco{s}^{-\frac{14}{3}}xco{s}^{4}xse{c}^{4}xdx$

$\Rightarrow I=\int \frac{si{n}^{\frac{2}{3}}x}{co{s}^{\frac{2}{3}}x}se{c}^{4}xdx$

$\Rightarrow I=\int ta{n}^{\frac{2}{3}}x\left(1+ta{n}^{2}x\right)se{c}^{2}xdx$

Put $tanx=t\Rightarrow se{c}^{2}xdx=dt$

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

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