3.6 Q-18

Question Statement

Evaluate the integral:

$$\int_{0}^{\frac{\pi}{4}} \cos^4 t , dt$$

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Background and Explanation

To solve this integral, we need to use trigonometric identities and techniques such as expanding powers of cosine and applying standard integrals for trigonometric functions. A key identity that can simplify this problem is the double angle formula and further expansion of trigonometric powers.

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Solution

We begin by expanding the given integral. We use the identity for $\cos^4 t$ and express it as:

$$\cos^4 t = \left( 2 \cos^2 t \right) \left( 2 \cos^2 t \right)$$

Now, we apply the identity $\cos^2 t = \frac{1 + \cos 2t}{2}$ to each term:

$$= \frac{1}{4} \int_{0}^{\frac{\pi}{4}} (1 + \cos 2t)^2 , dt$$

Now, expand the square:

$$= \frac{1}{4} \int_{0}^{\frac{\pi}{4}} \left( 1 + 2 \cos 2t + \cos^2 2t \right) , dt$$

We can now split the integral into three parts:

$$= \frac{1}{4} \left[ \int_{0}^{\frac{\pi}{4}} 1 , dt + 2 \int_{0}^{\frac{\pi}{4}} \cos 2t , dt + \int_{0}^{\frac{\pi}{4}} \cos^2 2t , dt \right]$$

For the second and third terms, use the identity for $\cos^2 x = \frac{1 + \cos 2x}{2}$ to handle $\cos^2 2t$:

$$= \frac{1}{4} \left[ \int_{0}^{\frac{\pi}{4}} 1 , dt + 2 \int_{0}^{\frac{\pi}{4}} \cos 2t , dt + \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \left( 1 + \cos 4t \right) , dt \right]$$

Now, we compute each integral individually.

1. Integral of 1:

$$\int_{0}^{\frac{\pi}{4}} 1 , dt = \frac{\pi}{4}$$

2. Integral of $\cos 2t$:

$$\int_{0}^{\frac{\pi}{4}} \cos 2t , dt = \frac{\sin 2t}{2} \Big|_0^{\frac{\pi}{4}} = \frac{\sin \frac{\pi}{2}}{2} - \frac{\sin 0}{2} = \frac{1}{2}$$

3. Integral of $\cos^2 2t$:

For $\cos^2 2t$, we use the identity and break it down:

$$\int_{0}^{\frac{\pi}{4}} \cos^2 2t , dt = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (1 + \cos 4t) , dt$$

Now, split this integral:

• Integral of 1:

$$\int_{0}^{\frac{\pi}{4}} 1 , dt = \frac{\pi}{4}$$

• Integral of $\cos 4t$:

$$\int_{0}^{\frac{\pi}{4}} \cos 4t , dt = \frac{\sin 4t}{4} \Big|_0^{\frac{\pi}{4}} = \frac{\sin \pi}{4} - \frac{\sin 0}{4} = 0$$

Thus:

$$\frac{1}{2} \left( \frac{\pi}{4} + 0 \right) = \frac{\pi}{8}$$

Now, putting everything together:

$$\frac{1}{4} \left[ \frac{\pi}{4} + 2 \times \frac{1}{2} + \frac{\pi}{8} \right]$$

Simplify:

$$= \frac{1}{4} \left[ \frac{\pi}{4} + \frac{\pi}{4} + \frac{\pi}{8} + 1 \right] = \frac{1}{4} \left[ \frac{2\pi}{4} + \frac{\pi}{8} + 1 \right] = \frac{1}{4} \left[ \frac{3\pi}{8} + 1 \right]$$

Finally:

$$= \frac{3\pi + 8}{32}$$

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Key Formulas or Methods Used

• Trigonometric identity: $\cos^2 t = \frac{1 + \cos 2t}{2}$
• Standard integrals:
  • $\int \cos x , dx = \sin x$
  • $\int 1 , dx = x$

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Summary of Steps

1. Expand $\cos^4 t$ using trigonometric identities.
2. Break the integral into simpler parts.
3. Use trigonometric identities for $\cos^2 t$ and $\cos^2 2t$.
4. Compute each integral separately.
5. Combine the results and simplify to get the final answer:

$$\frac{3\pi + 8}{32}$$