3.6 Q-20

Question Statement

Evaluate the integral:

$$\int_{0}^{\frac{\pi}{4}}\left(1 + \cos^2 \theta\right) \tan^2 \theta , d\theta$$

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Background and Explanation

This problem involves integrating a trigonometric expression with both $\tan^2 \theta$ and $\cos^2 \theta$. To solve it, we can break the integral into simpler components by applying trigonometric identities and methods such as the secant identity and the double angle identity. This will allow us to reduce the complexity of the integral step by step.

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Solution

We will break down the integral into manageable parts:

1. Split the integral: Start by splitting the original integral into two parts:

$$\int_{0}^{\frac{\pi}{4}} \left(1 + \cos^2 \theta \right) \tan^2 \theta , d\theta$$

This becomes:

$$\int_{0}^{\frac{\pi}{4}} \tan^2 \theta , d\theta + \int_{0}^{\frac{\pi}{4}} \cos^2 \theta \cdot \frac{\sin^2 \theta}{\cos^2 \theta} , d\theta$$

2. Simplify using identities: Using the identity $\tan^2 \theta = \sec^2 \theta - 1$, the second integral becomes:

$$\int_{0}^{\frac{\pi}{4}} (\sec^2 \theta - 1) , d\theta + \int_{0}^{\frac{\pi}{4}} \sin^2 \theta , d\theta$$

Now, the integral is split into three parts:

$$\int_{0}^{\frac{\pi}{4}} \sec^2 \theta , d\theta - \int_{0}^{\frac{\pi}{4}} 1 , d\theta + \frac{1}{2} \int_{0}^{\frac{\pi}{4}} 2 \sin^2 \theta , d\theta$$

3. Solve the first integral: The integral of $\sec^2 \theta$ is straightforward:

$$\int \sec^2 \theta , d\theta = \tan \theta$$

So we have:

$$\tan \theta \Big|_0^{\frac{\pi}{4}}$$

Evaluating:

$$\tan \frac{\pi}{4} - \tan 0 = 1 - 0 = 1$$

4. Solve the second integral: The integral of $1$ is just:

$$\int 1 , d\theta = \theta$$

So:

$$\theta \Big|_0^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$

5. Solve the third integral: The third part requires using the identity for $\sin^2 \theta$, which is:

$$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$

Substituting this in:

$$\frac{1}{2} \int_{0}^{\frac{\pi}{4}} (1 - \cos 2\theta) , d\theta$$

This splits into two simpler integrals:

$$\frac{1}{2} \left( \int_{0}^{\frac{\pi}{4}} 1 , d\theta - \int_{0}^{\frac{\pi}{4}} \cos 2\theta , d\theta \right)$$

The first integral is:

$$\int 1 , d\theta = \theta \Big|_0^{\frac{\pi}{4}} = \frac{\pi}{4}$$

The second integral requires the antiderivative of $\cos 2\theta$:

$$\int \cos 2\theta , d\theta = \frac{\sin 2\theta}{2}$$

So:

$$\frac{1}{2} \left( \frac{\pi}{4} - \frac{\sin 2\theta}{2} \Big|_0^{\frac{\pi}{4}} \right)$$

Evaluating $\sin 2\theta$ at the limits:

$$\sin 2 \times \frac{\pi}{4} = \sin \frac{\pi}{2} = 1 \quad \text{and} \quad \sin 0 = 0$$

Therefore:

$$\frac{1}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right)$$

Simplifying:

$$= \frac{1}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{1}{2} \times \frac{\pi - 2}{4} = \frac{\pi - 2}{8}$$

6. Final result: Now, combine all the results:

$$1 - \frac{\pi}{4} + \frac{\pi - 2}{8}$$

Simplifying:

$$= 1 - \frac{\pi}{4} + \frac{\pi}{8} - \frac{2}{8} = 1 - \frac{\pi}{8} - \frac{2}{8} = \frac{3}{4} - \frac{\pi}{8}$$

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Key Formulas or Methods Used

• Identity for tangent: $\tan^2 \theta = \sec^2 \theta - 1$
• Double angle identity: $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$
• Standard integrals:
  • $\int \sec^2 \theta , d\theta = \tan \theta$
  • $\int \cos 2\theta , d\theta = \frac{\sin 2\theta}{2}$
  • $\int 1 , d\theta = \theta$

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Summary of Steps

1. Split the integral into two parts using trigonometric identities.
2. Solve $\int \sec^2 \theta , d\theta$, $\int 1 , d\theta$, and $\int \sin^2 \theta , d\theta$.
3. Simplify the integrals using known formulas and evaluate the limits.
4. Combine the results to find the final value:

$$\frac{3}{4} - \frac{\pi}{8}$$