Question

The value of \( \int_{0}^{20\pi} \left( \sin^4 x + \cos^4 x \right)\, dx \) is equal to

• A. \( 15\pi \)
• B. \( \frac{15\pi}{2} \)
• C. \( 25\pi \)
• D. \( \frac{25\pi}{2} \)

Hint:

For expressions like \( \sin^4 x+\cos^4 x \), first reduce powers using identities. A very useful identity is \( \sin^4 x+\cos^4 x=1-2\sin^2 x \cos^2 x \), which makes integration much easier.

Answer 1

The Correct Option is A

Step 1: Simplify the integrand.
We use the identity:
\[ \sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2 x \cos^2 x \] Since \( \sin^2 x+\cos^2 x=1 \), we get:
\[ \sin^4 x+\cos^4 x=1-2\sin^2 x \cos^2 x \] Now, using \( \sin 2x=2\sin x \cos x \), we have:
\[ \sin^2 x \cos^2 x=\frac{1}{4}\sin^2 2x \] So,
\[ \sin^4 x+\cos^4 x=1-\frac{1}{2}\sin^2 2x \]
Step 2: Further simplify using the identity for \( \sin^2 2x \).
We know that:
\[ \sin^2 2x=\frac{1-\cos 4x}{2} \] Substituting this,
\[ \sin^4 x+\cos^4 x=1-\frac{1}{2}\cdot \frac{1-\cos 4x}{2} \] \[ =1-\frac{1-\cos 4x}{4} \] \[ =\frac{3+\cos 4x}{4} \] Thus, the integral becomes:
\[ \int_{0}^{20\pi} \left( \sin^4 x+\cos^4 x \right)\,dx =\int_{0}^{20\pi}\frac{3+\cos 4x}{4}\,dx \]
Step 3: Integrate term by term.
\[ \int_{0}^{20\pi}\frac{3+\cos 4x}{4}\,dx =\frac{1}{4}\int_{0}^{20\pi}3\,dx+\frac{1}{4}\int_{0}^{20\pi}\cos 4x\,dx \] Now,
\[ \frac{1}{4}\int_{0}^{20\pi}3\,dx=\frac{3}{4}\left[ x \right]_{0}^{20\pi} =\frac{3}{4}(20\pi)=15\pi \] Also,
\[ \frac{1}{4}\int_{0}^{20\pi}\cos 4x\,dx =\frac{1}{4}\left[\frac{\sin 4x}{4}\right]_{0}^{20\pi} \] \[ =\frac{1}{16}\left(\sin 80\pi-\sin 0\right)=0 \] Hence, total value is:
\[ 15\pi+0=15\pi \]
Step 4: Conclusion.
Therefore, the value of the integral is \( 15\pi \).

Final Answer: \( 15\pi \)