We are given the integral: \[ \int \frac{\sec^2(\sqrt{2x+5})}{\sqrt{2x+5}} \, dx \] To solve this, let's substitute \( u = \sqrt{2x + 5} \). This implies: \[ du = \frac{1}{2\sqrt{2x + 5}} \cdot 2 \, dx = \frac{1}{\sqrt{2x+5}} \, dx \]
Thus, the integral becomes: \[ \int \sec^2(u) \, du = \tan(u) + C \] Now, substituting \( u = \sqrt{2x+5} \), we get: \[ \tan(\sqrt{2x+5}) + C \] The correct answer is \( \frac{1}{2} \log |2x + 5| \).
If, \( I_n = \int_{-\pi}^{\pi} \frac{\cos(nx)(1+2^x)}{dx} \), where \( n = 0, 1, 2, \dots \), then which of the following are correct?
A. \( I_n = I_{n+2} \), for all \( n = 0, 1, 2, \dots \)
B. \( I_n = 0 \), for all \( n = 0, 1, 2, \dots \)
C. \( \sum_{n=1}^{10} I_n = 2^{10} \)
D. \( \sum_{n=1}^{10} I_n = 0 \)