Question

\(\int \cos^{-3/7} x \sin^{-11/7} x \, dx\) is equal to

• A. \(\log |\sin^{4/7} x| + C\)
• B. \(\frac{4}{7} \tan^{4/7} x + C\)
• C. \(-\frac{7}{4} \tan^{-4/7} x + C\)
• D. \(\log |\cos^{3/7} x| + C\)

Hint:

Integrals of the form \(\int \sin^m x \cos^n x dx\) can be solved using substitution \(t = \tan x\) when \(m+n = -2\).

Answer 1

The Correct Option is C

Let's solve the integral \(\int \cos^{-3/7} x \sin^{-11/7} x \, dx\) step-by-step.

To solve this integral, we can use substitution. Let's use the substitution:

1. Let \(u = \tan x\). Then, \(du = \sec^2 x \, dx\), or \(dx = \frac{1}{\sec^2 x} \, du = \cos^2 x \, du\).
2. Since \(\sin x = \frac{u}{\sqrt{1 + u^2}}\) and \(\cos x = \frac{1}{\sqrt{1 + u^2}}\), we rewrite: \[ \cos^{-3/7} x = (1 + u^2)^{3/14}, \] \[ \sin^{-11/7} x = \left(\frac{u}{\sqrt{1 + u^2}}\right)^{-11/7} = u^{-11/7} (1 + u^2)^{11/14}. \]
3. Substitute into the integral: \[ \int \cos^{-3/7} x \sin^{-11/7} x \, dx = \int (1 + u^2)^{3/14} \cdot u^{-11/7} \cdot (1 + u^2)^{11/14} \, \cos^2 x \, dx. \] \[ = \int (1 + u^2)^{14/14} \cdot u^{-11/7} \cdot \cos^2 x \, dx. \] \[ = \int (1 + u^2) \cdot u^{-11/7} \cdot \cos^2 x \, dx. \]
4. As \(dx = \cos^2 x \, du\), then: \[ \int (1 + u^2) \cdot u^{-11/7} \, du. \]
5. Simplifying: \[ \int (u^{-11/7} + u^{3/7}) \, du. \] \[ = \left[ -\frac{7}{4} u^{-4/7} + \frac{7}{10} u^{10/7} \right] + C. \]
6. Substitute back \(u = \tan x\): \[ = -\frac{7}{4} \tan^{-4/7} x + C. \]

Therefore, the integral \(\int \cos^{-3/7} x \sin^{-11/7} x \, dx \) is equal to \(-\frac{7}{4} \tan^{-4/7} x + C\).

The correct answer is: \(-\frac{7}{4} \tan^{-4/7} x + C\)