Question

If \( I_1 = \int \sin^6 x \, dx \) and \( I_2 = \int \cos^6 x \, dx \) then \( I_1 + I_2 = \)

• A. \( \frac{5x}{8} + \frac{3\cos 4x}{32} + c \)
• B. \( \frac{1}{32} (20x - 3\sin 4x) + c \)
• C. \( \frac{1}{32} (20x + 3\sin 4x) + c \)
• D. \( \frac{5x}{4} + \frac{3\sin 4x}{16} + c \)

Hint:

The identities for sums of powers of sine and cosine are very useful. Remember \( \sin^4x+\cos^4x = 1 - 2\sin^2x\cos^2x \) and \( \sin^6x+\cos^6x = 1 - 3\sin^2x\cos^2x \). These allow quick simplification before integration.

Answer 1

The Correct Option is C

We want to evaluate \( I_1 + I_2 = \int (\sin^6 x + \cos^6 x) \, dx \).
Let's simplify the integrand \( \sin^6 x + \cos^6 x \).
We can use the identity \( a^3+b^3 = (a+b)(a^2-ab+b^2) \). Let \( a=\sin^2 x \) and \( b=\cos^2 x \).
\( (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \).
Since \( \sin^2 x + \cos^2 x = 1 \), this simplifies to:
\( (\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) \).
We can rewrite \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \).
So the expression becomes \( (1 - 2\sin^2 x \cos^2 x) - \sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x \).
Now we use the double angle identity \( \sin(2x) = 2\sin x \cos x \), so \( \sin x \cos x = \frac{1}{2}\sin(2x) \).
\( \sin^2 x \cos^2 x = (\frac{1}{2}\sin(2x))^2 = \frac{1}{4}\sin^2(2x) \).
The integrand is \( 1 - 3 \cdot \frac{1}{4}\sin^2(2x) = 1 - \frac{3}{4}\sin^2(2x) \).
To integrate this, use the power reduction formula \( \sin^2\theta = \frac{1-\cos(2\theta)}{2} \).
\( 1 - \frac{3}{4} \left( \frac{1-\cos(4x)}{2} \right) = 1 - \frac{3}{8}(1-\cos(4x)) = 1 - \frac{3}{8} + \frac{3}{8}\cos(4x) = \frac{5}{8} + \frac{3}{8}\cos(4x) \).
Now, we integrate this expression.
\( \int (\frac{5}{8} + \frac{3}{8}\cos(4x)) \, dx = \frac{5}{8}x + \frac{3}{8} \cdot \frac{\sin(4x)}{4} + c \).
\( = \frac{5x}{8} + \frac{3\sin(4x)}{32} + c \).
To match the form of option (C), we find a common denominator of 32.
\( = \frac{20x}{32} + \frac{3\sin(4x)}{32} + c = \frac{1}{32}(20x + 3\sin 4x) + c \).