Question

If \( u(n) = \int_0^{\frac{\pi}{2}} (1 + \sin t)^n \sin 2t \, dt,\; n \in \mathbb{N} \), then \( u(4) = \)

• A. \(\frac{28\pi}{5}\)
• B. \(\frac{128}{35}\)
• C. \(\frac{129}{15}\)
• D. \(\frac{68\pi}{15}\)

Hint:

When an integral involves powers of \(1 + \sin t\) and \(\sin 2t\), use trigonometric identities and substitutions like \(x = \sin t\) to simplify.

Answer 1

The Correct Option is C

We are given: \[ u(n) = \int_0^{\frac{\pi}{2}} (1 + \sin t)^n \cdot \sin 2t \, dt \] Use the identity: \[ \sin 2t = 2 \sin t \cos t \Rightarrow u(n) = \int_0^{\frac{\pi}{2}} (1 + \sin t)^n \cdot 2 \sin t \cos t \, dt \] Let \( I = 2 \int_0^{\frac{\pi}{2}} (1 + \sin t)^n \sin t \cos t \, dt \) Now, substitute \( x = \sin t \), so that: \[ dx = \cos t \, dt \quad \text{and when } t = 0,\ x = 0;\ t = \frac{\pi}{2},\ x = 1 \] Thus, \[ I = 2 \int_0^1 (1 + x)^n x \, dx = 2 \int_0^1 (1 + x)^n x \, dx \] We now compute \( u(4) \Rightarrow n = 4 \): So, \[ u(4) = 2 \int_0^1 (1 + x)^4 x \, dx \] Expand: \[ (1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 \Rightarrow (1 + x)^4 x = x + 4x^2 + 6x^3 + 4x^4 + x^5 \] Now integrate term by term: \[ \int_0^1 x \, dx = \frac{1}{2},\; \int_0^1 4x^2 \, dx = \frac{4}{3},\; \int_0^1 6x^3 \, dx = \frac{6}{4} = \frac{3}{2},\; \int_0^1 4x^4 \, dx = \frac{4}{5},\; \int_0^1 x^5 \, dx = \frac{1}{6} \] Add all: \[ \frac{1}{2} + \frac{4}{3} + \frac{3}{2} + \frac{4}{5} + \frac{1}{6} = \frac{129}{30} \Rightarrow u(4) = 2 \cdot \frac{129}{30} = \frac{129}{15} \] \[ \boxed{u(4) = \frac{129}{15}} \]