Question

\((\pi / 2) \int_{(\pi / 4)} \sqrt{(1-\sin 2 x) d x}\) is equal to \(\ldots \ldots \ldots\) (a) \(\sqrt{2}+1\) (b) \(\sqrt{2}-1\) (c) \(1-\sqrt{2}\) (d) 0

Short answer

\( \frac{\pi}{2} \cdot \frac{1 - \sqrt{2}}{2} = \frac{1}{2}({1 - \sqrt{2}})\) The answer is: \( (b) \: \sqrt{2}-1 \)

Step-by-step solution

Use the identity for sin(2x)

The identity for sin(2x) is: \[\sin{2x} = 2\sin{x}\cos{x}\] So, we will use this identity in the given expression: \(\frac{\pi}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 - \sin{2x}} dx = \frac{\pi}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 - 2\sin{x}\cos{x}} dx \)

Perform the substitution

Now, let's perform the substitution by setting: \[\cos{x} = t\] Then, \[-\sin{x} dx = dt\] Now, the integral becomes: \(-\frac{\pi}{2}\int_{\frac{\sqrt{2}}{2}}^{1} \sqrt{1 - 2t(1-t)} dt \)

Simplify the expression inside the square root

Now, let's simplify the expression inside the square root: \[\sqrt{1 - 2t(1-t)} = \sqrt{1 - 2t + 4t^2 - 2t} = \sqrt{4t^2 - 4t + 1}\] The integral becomes: \(-\frac{\pi}{2}\int_{\frac{\sqrt{2}}{2}}^{1} \sqrt{4t^2 - 4t + 1} dt \)

Simplification by completing the square

Notice that the expression inside the square root is a quadratic equation. We can rewrite it by completing the square as follows: \[\sqrt{4t^2 - 4t + 1} = \sqrt{(2t-1)^2} = |2t-1|\] Now, since t is between \(\frac{\sqrt{2}}{2}\) and 1, |2t - 1| simplifies to 1 - 2t. So, the integral becomes: \(-\frac{\pi}{2}\int_{\frac{\sqrt{2}}{2}}^{1} (1 - 2t) dt \)

Integrate and evaluate

Now, let's integrate the expression: \[-\frac{\pi}{2}\int_{\frac{\sqrt{2}}{2}}^{1} (1 - 2t) dt = -\frac{\pi}{2} \left[t - t^2\right]_{\frac{\sqrt{2}}{2}}^{1}\] Evaluate at the boundaries: \[-\frac{\pi}{2} \left[ \left(1 - 1^2\right) - \left(\frac{\sqrt{2}}{2} - \left(\frac{\sqrt{2}}{2}\right)^2\right) \right] = -\frac{\pi}{2} \left[ 0 - \left(\frac{\sqrt{2}}{2} - \frac{1}{2}\right) \right] = -\frac{\pi}{2}\left[\frac{1 - \sqrt{2}}{2}\right]\] Therefore, the value of the given integral is: \(\frac{\pi}{2} \cdot \frac{1 - \sqrt{2}}{2} = \frac{1}{2}({1 - \sqrt{2}})\) Comparing this result to the choices given, we find that the answer is: \( \boxed{(b) \: \sqrt{2}-1} \)

Key concepts

Definite Integral

In the realm of calculus, the definite integral is highly valued for its ability to quantify the area under a curve. Specifically, when we look at an expression like \(\int_a^b f(x) dx\) with limits from \(a\) to \(b\), what we're seeking is the total accumulation of \(f(x)\)'s values within this interval. In the context of JEE Maths problems, understanding how to compute the definite integral is essential for solving a broad range of questions involving areas, volumes, and average values.

When faced with a definite integral problem, such as the one given in the exercise, employing the correct approach is paramount. The first step often involves simplifying the integrand using algebraic manipulation or trigonometric identities. Once simplified, the integral is solved within the specified bounds \(a\) and \(b\), and the resulting values are substituted back to give the final answer. Careful attention to the limits of integration and any changes therein, especially when substitution methods are used, ensures accurate results.

Trigonometric Identities

Trigonometric identities play a pivotal role in simplifying complex expressions involving trigonometric functions. These identities are mathematical equations that relate one trigonometric function to another, allowing for the transformation of an integral's integrand into a more workable form. For JEE aspirants, mastering these identities—including those for sine, cosine, tangent, and their inverses—is vital.

In the context of the given exercise, the identity for \(\sin(2x) = 2\sin(x)\cos(x)\) is utilized to deconstruct the integrand's square root. This is a classic example of how recognizing and applying trigonometric identities can transform a seemingly complicated integral into one that's easier to handle. Gratifying the need for simplicity, these identities often set the stage for the next critical step in solving integrals: the substitution method.

Substitution Method

Utilized extensively in integrating functions, the substitution method—also known as u-substitution—is a technique that involves replacing part of the integrand with a new variable to simplify the integration process. This strategy can be likened to finding the 'right key' for a 'mathematical lock,' and when the correct substitution is made, the integral often becomes far more manageable.

In our exercise, the substitution \(\cos(x) = t\) converts the expression from a function of \(x\) to a function of \(t\). This not only simplifies the integrand but also changes the integration limits accordingly. The choice of substitution is crucial and largely depends on the form of the integrand and one's experience with similar problems. Once the integration is performed with respect to the new variable, don't forget to revert back to the original variable if necessary and apply the new limits to find the definite integral—thus unlocking the solution to the problem.