Question

In the following exercises, compute each definite integral. \(\int_{0}^{1 / 2} \frac{\sin \left(\tan ^{-1} t\right)}{1+t^{2}} d t\)

Short answer

The answer is \(1 - \frac{2\sqrt{5}}{5}\).

Step-by-step solution

Identify a Suitable Substitution

Examine the integral \( \int_{0}^{1/2} \frac{\sin(\tan^{-1} t)}{1+t^2} \, dt \). Notice that the derivative of \( \tan^{-1} t \) is related to the integrand: \( \frac{d}{dt}(\tan^{-1} t) = \frac{1}{1+t^2} \). This suggests using the substitution \( u = \tan^{-1} t \).

Compute the Differential

Differentiate \( u = \tan^{-1} t \) with respect to \( t \) to find \( du \). We have \( du = \frac{1}{1+t^2} \, dt \), or equivalently, \( dt = (1+t^2) \, du \). However, since \( dt = (1+t^2) \, du \), we can simplify: \( dt = du \) since the term \( \frac{1}{1+t^2} \, dt \) is exactly \( du \).

Change Limits of Integration

Substitute the limits of integration: when \( t=0 \), \( u = \tan^{-1}(0) = 0 \); when \( t=1/2 \), \( u = \tan^{-1}(1/2) \). This changes the limits from \([0, 1/2]\) in \( t \) to \([0, \tan^{-1}(1/2)]\) in \( u \).

Integrate with Respect to \( u \)

Recognize that the integral becomes simply \( \int_{0}^{\tan^{-1}(1/2)} \sin u \, du \). We know that the integral of \( \sin u \) is \(-\cos u \). Thus, \( \int \sin u \, du = -\cos u + C \).

Evaluate the Definite Integral

Evaluate \( -\cos u \) from \( u = 0 \) to \( u = \tan^{-1}(1/2) \). This gives: \([-\cos(\tan^{-1}(1/2))] - [-\cos(0)].\)Recall \( \cos(0) = 1 \). The angle \( \tan^{-1}(1/2) \) is represented by a right triangle where the opposite side is 1 and adjacent side is 2, hence \( \cos(\tan^{-1}(1/2)) = \frac{2}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{5}} \). Therefore, the expression simplifies to:\( -\frac{2}{\sqrt{5}} - (-1) = 1 - \frac{2}{\sqrt{5}} \).

Simplify the Result

The integral evaluates to \( 1 - \frac{2}{\sqrt{5}} \). To remove the square root from the denominator, multiply by \( \frac{\sqrt{5}}{\sqrt{5}} \) to get \( 1 - \frac{2\sqrt{5}}{5} \).

Key concepts

Substitution Method

The substitution method is a powerful technique used to simplify the integration process, especially when faced with complex integrals. By carefully choosing a new variable, we can transform the integral into a simpler form.
To apply it, we first identify a part of the integral that can be substituted with a new variable, say 'u'.

• In the given exercise, the variable substitution chosen was \( u = \tan^{-1}(t) \). This substitution is strategic because the derivative of \( \tan^{-1}(t) \) is \( \frac{1}{1+t^2} \), which appears in the integrand.
• Next, we compute the differential of 'u', finding \( du = \frac{1}{1+t^2} \, dt \). This transformation allows us to rewrite the original integral in terms of 'u'.
• By replacing both the function in the integrand and the differential \( dt \) with \( du \), the integral becomes much easier to handle.

The substitution method essentially "translates" complex integrals into simpler ones that are more straightforward to solve. This method is particularly useful when the integrand and its differential neatly align, enabling a direct substitution.

Trigonometric Functions

Trigonometric functions are fundamental in calculus, often appearing in integration problems due to their wave-like properties and periodicity. In this exercise, the function involved is \( \sin(u) \), which is a classic trigonometric function.

• The integral \( \int \sin(u) \, du \) appears straightforward, thanks to the substitution method. Calculating this integral leads to \( -\cos(u) + C \), a standard result.
• In the context of definite integrals, like our exercise, we evaluate \( -\cos(u) \) between specified limits. The periodical nature of cosine and sine makes trigonometric functions unique in taking specific values over defined intervals.

It's essential to recognize and manipulate these functions' properties to simplify the integration process effectively. Understanding how to integrate these functions, especially after substitution, is a key skill in tackling more complicated calculus problems.

Limits of Integration

Transitioning from indefinite to definite integrals requires careful consideration of limits. The limits of integration define the range over which the function is integrated, turning the problem from one of generality to specificity.

• Initially, our integral has limits from \( t = 0 \) to \( t = \frac{1}{2} \). With the substitution \( u = \tan^{-1}(t) \), these limits transform.
• When \( t = 0 \), \( u = \tan^{-1}(0) = 0 \). For \( t = \frac{1}{2} \), \( u = \tan^{-1}(\frac{1}{2}) \), which shifts the original limits to \( [0, \tan^{-1}(\frac{1}{2})] \).
• These new boundaries are critical for the integration process as they ensure the definite integral is evaluated correctly, providing a precise numerical result. These transformed limits are directly substituted into \( \int \sin(u) \, du \), leading to an exact evaluation: \( 1 - \frac{2}{\sqrt{5}} \), which simplifies further.

Correctly changing integration limits is essential when employing substitution, as it ensures the integral remains equivalent in its new form.