Question

\(\int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin \theta}\cos^3\theta\ d\theta\) is equal to

• A. \(\frac{8}{23}\)
• B. \(\frac{8}{21}\)
• C. \(\frac{7}{23}\)
• D. \(\frac{7}{21}\)

Answer 1

The Correct Option is B

We evaluate the integral: \[ \int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin \theta}\cos^3\theta\ d\theta \] 

Step 1: Use substitution
Let \(\sin \theta = t \Rightarrow \cos \theta\, d\theta = dt\) When \(\theta = 0\), \(t = 0\)
When \(\theta = \frac{\pi}{2}\), \(t = 1\) We know: \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - t^2 \Rightarrow \cos \theta = \sqrt{1 - t^2} \] But this approach is complicated. Try different substitution.

 Step 2: Let us try \( \theta = \sin^{-1}x \Rightarrow x = \sin\theta \)
Then, \( d\theta = \frac{1}{\sqrt{1 - x^2}} dx \), and \( \cos\theta = \sqrt{1 - x^2} \) So: \[ \int_0^{\frac{\pi}{2}} \sqrt{\sin \theta} \cos^3 \theta\, d\theta = \int_0^1 \sqrt{x} (1 - x^2)^{3/2} \cdot \frac{1}{\sqrt{1 - x^2}} dx \] Simplify: \[ = \int_0^1 \sqrt{x} (1 - x^2) dx = \int_0^1 x^{1/2}(1 - x^2)\ dx \] 

Step 3: Expand and integrate
\[ \int_0^1 x^{1/2} - x^{5/2} dx = \int_0^1 x^{1/2} dx - \int_0^1 x^{5/2} dx \] \[ = \left[\frac{x^{3/2}}{3/2}\right]_0^1 - \left[\frac{x^{7/2}}{7/2}\right]_0^1 = \frac{2}{3} - \frac{2}{7} = \frac{14 - 6}{21} = \frac{8}{21} \] 

Final answer: \(\frac{8}{21}\)

Answer 2

To solve the integral \(\int\limits_{0}^{\frac{\pi}{2}} \sqrt{\sin \theta} \cos^3 \theta \, d\theta\), we can use a substitution method. Let's set \(u = \sin \theta\). Then, \(du = \cos \theta \, d\theta\).

When \(\theta = 0\), \(u = \sin 0 = 0\).
When \(\theta = \frac{\pi}{2}\), \(u = \sin \frac{\pi}{2} = 1\).

The integral becomes:

\[ \int\limits_{0}^{1} \sqrt{u} (1 - u^2) \, du \]

Now, we can split the integral into two parts:

\[ \int\limits_{0}^{1} u^{1/2} \, du - \int\limits_{0}^{1} u^{5/2} \, du \]

Evaluating each part separately:

\[ \int\limits_{0}^{1} u^{1/2} \, du = \left[ \frac{2}{3} u^{3/2} \right]_{0}^{1} = \frac{2}{3} \]

\[ \int\limits_{0}^{1} u^{5/2} \, du = \left[ \frac{2}{7} u^{7/2} \right]_{0}^{1} = \frac{2}{7} \]

Combining these results:

\[ \frac{2}{3} - \frac{2}{7} = \frac{14}{21} - \frac{6}{21} = \frac{8}{21} \]

Therefore, the value of the integral is:

\[ \boxed{\frac{8}{21}} \]