Question

\(\int\sqrt{\cosec x-\sin x}\ dx=\)

• A. \(2\sqrt{\sin x}+C\)
• B. \(\sqrt{\sin x}+C\)
• C. \(\frac{2}{\sqrt{\sin x}}+C\)
• D. \(\frac{\sqrt{\sin x}}{2}+C\)

Hint:

When working with integrals involving trigonometric functions, try simplifying the integrand by applying trigonometric identities or substitutions. Recognizing standard integral forms early in the process can greatly speed up the solution, as demonstrated in this example.

Answer 1

The Correct Option is A

\(\int \sqrt{\csc x - \sin x} \, dx\)

Step 1: Analyze the integrand

We begin by analyzing the integrand \(\sqrt{\csc x - \sin x}\). The expression involves trigonometric functions \(\csc x\) and \(\sin x\), which suggests there might be an identity or simplification that can be applied to make the integration easier.

Step 2: Simplify the integrand

We rewrite \(\csc x\) as \(\frac{1}{\sin x}\), giving us the expression \(\sqrt{\frac{1}{\sin x} - \sin x}\). This form is easier to handle and suggests a connection with a standard trigonometric identity.

Step 3: Further simplify the expression

To simplify further, we combine the terms inside the square root: \(\sqrt{\frac{1 - \sin^2 x}{\sin x}}\). Recognizing that \(1 - \sin^2 x = \cos^2 x\), we obtain the expression \(\sqrt{\frac{\cos^2 x}{\sin x}} = \frac{\cos x}{\sqrt{\sin x}}\).

Step 4: Perform the integration

Now we integrate \(\int \frac{\cos x}{\sqrt{\sin x}} \, dx\). We use the substitution \(u = \sin x\), which gives \(du = \cos x \, dx\). The integral becomes \(\int \frac{1}{\sqrt{u}} \, du\), which is a standard integral.

Step 5: Solve the integral

The integral \(\int \frac{1}{\sqrt{u}} \, du\) is straightforward to evaluate, yielding \(2\sqrt{u} + C = 2\sqrt{\sin x} + C\).

Conclusion:
The correct answer is (A): \(2\sqrt{\sin x} + C\).