Question:hard

The value of the definite integral $\int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx$ is:

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For integrals of the form $\int_0^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} \, dx$, the answer is always half the upper limit, which is $\frac{\pi}{4}$.
Updated On: Jun 3, 2026
  • $\frac{\pi}{4}$
  • $\frac{\pi}{2}$
  • $\pi$
  • $0$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Recall the swap property.
A very useful rule says we can replace $x$ with $(a + b - x)$ in a definite integral without changing its value. For limits $0$ to $\frac{\pi}{2}$, that means replacing $x$ with $\frac{\pi}{2} - x$. This swaps sine and cosine.

Step 2: Name the integral.
Call the integral $I$.
\[ I = \int_0^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x}\,dx \]

Step 3: Apply the swap.
Replace $x$ with $\frac{\pi}{2} - x$. Then $\sin x$ turns into $\cos x$ and the other way round.
\[ I = \int_0^{\pi/2}\frac{\cos^{3/2}x}{\cos^{3/2}x + \sin^{3/2}x}\,dx \]

Step 4: Add the two forms.
Add the first form of $I$ and this new form. The tops add to give the same bottom.
\[ 2I = \int_0^{\pi/2}\frac{\sin^{3/2}x + \cos^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x}\,dx \]

Step 5: Simplify the fraction.
The top and bottom are equal, so the fraction is just $1$.
\[ 2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2} \]

Step 6: Solve for $I$.
Divide both sides by $2$.
\[ I = \frac{\pi}{4} \]
\[ \boxed{\dfrac{\pi}{4}} \]
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