Question

The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \, \csc^{\frac{4}{3}} x \, dx$ is equal to

• A. $3^{\frac{5}{6}}-3^{\frac{2}{3}}$
• B. $3^{\frac{7}{6}}-3^{\frac{5}{6}}$
• C. $3^{\frac{5}{3}}-3^{\frac{1}{3}}$
• D. $3^{\frac{4}{3}}-3^{\frac{1}{3}}$

Hint:

Calculus Tip: For integrals of the form $\int \sin^m x \cos^n x dx$, if $(m+n)$ is a negative even integer (here, $-4/3 - 2/3 = -2$), dividing the numerator and denominator by $\cos^{-(m+n)}x$ (here, $\cos^2 x$) is the standard technique to convert the expression into terms of $\tan x$ and $\sec^2 x$.

Answer 1

The Correct Option is B

Concept: Calculus - Definite Integration using Substitution (converting to $\tan x$ and $\sec^2 x$).

Step 1: Convert the secant and cosecant terms to sine and cosine. Rewrite the integral $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sec^{\frac{2}{3}}x~\csc^{\frac{4}{3}}x~dx$ using basic trigonometric ratios: $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\cos^{\frac{2}{3}}x \cdot \sin^{\frac{4}{3}}x}$.

Step 2: Manipulate the denominator to create a $\tan x$ term. To integrate expressions with fractional powers of sine and cosine, we aim to form $\tan x$ and $\sec^2 x$. Multiply and divide the denominator by $\cos^{\frac{4}{3}}x$: $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\left(\frac{\sin^{\frac{4}{3}}x}{\cos^{\frac{4}{3}}x}\right) \cdot \cos^{\frac{4}{3}}x \cdot \cos^{\frac{2}{3}}x}$. This simplifies to: $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\tan^{\frac{4}{3}}x \cdot \cos^{2}x}$. Since $\frac{1}{\cos^2 x} = \sec^2 x$, the integral becomes $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\tan^{\frac{4}{3}}x} dx$.

Step 3: Apply integration by substitution. Let $t = \tan x$. Then, differentiating both sides gives $dt = \sec^2 x~dx$.

Step 4: Update the limits of integration. When $x = \frac{\pi}{6}$, $t = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} = 3^{-\frac{1}{2}}$. When $x = \frac{\pi}{3}$, $t = \tan(\frac{\pi}{3}) = \sqrt{3} = 3^{\frac{1}{2}}$. The new integral is $I = \int_{3^{-1/2}}^{3^{1/2}} \frac{dt}{t^{4/3}} = \int_{3^{-1/2}}^{3^{1/2}} t^{-\frac{4}{3}} dt$.

Step 5: Integrate and evaluate the definite integral. Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1}$, we get: $\int t^{-\frac{4}{3}} dt = \frac{t^{-1/3}}{-1/3} = -3t^{-\frac{1}{3}}$. Now, apply the limits: $I = -3 \left[ t^{-\frac{1}{3}} \right]_{3^{-1/2}}^{3^{1/2}} = -3 \left[ (3^{1/2})^{-1/3} - (3^{-1/2})^{-1/3} \right]$. Simplify the exponents: $I = -3 \left[ 3^{-\frac{1}{6}} - 3^{\frac{1}{6}} \right] = -3 \cdot 3^{-\frac{1}{6}} + 3 \cdot 3^{\frac{1}{6}}$. Using exponent rules ($a^m \cdot a^n = a^{m+n}$): $I = -3^{1-\frac{1}{6}} + 3^{1+\frac{1}{6}} = -3^{\frac{5}{6}} + 3^{\frac{7}{6}} = 3^{\frac{7}{6}} - 3^{\frac{5}{6}}$. $$ \therefore \text{The integral evaluates to } 3^{\frac{7}{6}}-3^{\frac{5}{6}}. $$