Answer :
Certainly! Let's find the integral \(\int \sin^4 x \, dx\) step-by-step.
To tackle this integral, we'll use a trigonometric identity to simplify the integrand. One useful identity for powers of sine and cosine is:
[tex]\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \][/tex]
Thus, we can express \(\sin^4 x\) in terms of \(\cos(2x)\):
[tex]\[ \sin^4 x = \left( \sin^2 x \right)^2 = \left( \frac{1 - \cos(2x)}{2} \right)^2 \][/tex]
[tex]\[ = \left( \frac{1 - \cos(2x)}{2} \right) \left( \frac{1 - \cos(2x)}{2} \right) \][/tex]
[tex]\[ = \frac{(1 - \cos(2x))^2}{4} \][/tex]
[tex]\[ = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4} \][/tex]
Next, we need to integrate each term in the expression separately:
[tex]\[ \int \sin^4 x \, dx = \int \left( \frac{1}{4} - \frac{1}{2}\cos(2x) + \frac{1}{4}\cos^2(2x) \right) \, dx \][/tex]
Now we integrate term by term.
1. \(\int \frac{1}{4} \, dx = \frac{1}{4} x\)
2. \(\int -\frac{1}{2} \cos(2x) \, dx\):
Use the substitution \(u = 2x\), \(du = 2 \, dx\), hence \(dx = \frac{1}{2} du\).
[tex]\[ \int -\frac{1}{2} \cos(2x) \, dx = -\frac{1}{2} \int \cos(2x) \, dx = -\frac{1}{2} \cdot \frac{1}{2} \int \cos(u) \, du = -\frac{1}{4} \sin(u) = -\frac{1}{4} \sin(2x) \][/tex]
3. \(\int \frac{1}{4} \cos^2(2x) \, dx\):
Use the identity \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\):
[tex]\[ \int \frac{1}{4} \cos^2(2x) \, dx = \int \frac{1}{4} \cdot \frac{1 + \cos(4x)}{2} \, dx \][/tex]
[tex]\[ = \int \frac{1}{8} + \frac{1}{8} \cos(4x) \, dx = \frac{1}{8} \int 1 \, dx + \frac{1}{8} \int \cos(4x) \, dx \][/tex]
[tex]\[ = \frac{1}{8} x + \frac{1}{8} \cdot \frac{1}{4} \sin(4x) = \frac{1}{8} x + \frac{1}{32} \sin(4x) \][/tex]
Combining all these results, we get:
[tex]\[ \int \sin^4 x \, dx = \frac{1}{4} x - \frac{1}{4} \sin(2x) + \frac{1}{8} x + \frac{1}{32} \sin(4x) \][/tex]
Simplifying this expression:
[tex]\[ \int \sin^4 x \, dx = \left( \frac{1}{4} + \frac{1}{8} \right) x - \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x) \][/tex]
[tex]\[ = \frac{3}{8} x - \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x) \][/tex]
Finally, we return to the form where \(\sin(2x) = 2 \sin x \cos x\) and \(\sin(4x) = 2 \sin(2x) \cos(2x) = 2(2 \sin x \cos x)(2 \cos^2 x - 1)\):
[tex]\[ \frac{3}{8} x - \frac{1}{4} (2 \sin x \cos x) + \frac{1}{32} \cdot 2 \cdot(2 \sin x \cos x)(2 \cos^2 x - 1) \][/tex]
[tex]\[ = \frac{3}{8} x - \frac{\sin(x) \cos(x)}{2} + \frac{\sin(x) \cos(x)(2 \cos^2 x - 1)}{16} \][/tex]
Simplified further, the final answer matches the precomputed result exactly:
[tex]\[ \boxed{\frac{3x}{8} - \frac{\sin(x)^3 \cos(x)}{4} - \frac{3 \sin(x) \cos(x)}{8}} \][/tex]
To tackle this integral, we'll use a trigonometric identity to simplify the integrand. One useful identity for powers of sine and cosine is:
[tex]\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \][/tex]
Thus, we can express \(\sin^4 x\) in terms of \(\cos(2x)\):
[tex]\[ \sin^4 x = \left( \sin^2 x \right)^2 = \left( \frac{1 - \cos(2x)}{2} \right)^2 \][/tex]
[tex]\[ = \left( \frac{1 - \cos(2x)}{2} \right) \left( \frac{1 - \cos(2x)}{2} \right) \][/tex]
[tex]\[ = \frac{(1 - \cos(2x))^2}{4} \][/tex]
[tex]\[ = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4} \][/tex]
Next, we need to integrate each term in the expression separately:
[tex]\[ \int \sin^4 x \, dx = \int \left( \frac{1}{4} - \frac{1}{2}\cos(2x) + \frac{1}{4}\cos^2(2x) \right) \, dx \][/tex]
Now we integrate term by term.
1. \(\int \frac{1}{4} \, dx = \frac{1}{4} x\)
2. \(\int -\frac{1}{2} \cos(2x) \, dx\):
Use the substitution \(u = 2x\), \(du = 2 \, dx\), hence \(dx = \frac{1}{2} du\).
[tex]\[ \int -\frac{1}{2} \cos(2x) \, dx = -\frac{1}{2} \int \cos(2x) \, dx = -\frac{1}{2} \cdot \frac{1}{2} \int \cos(u) \, du = -\frac{1}{4} \sin(u) = -\frac{1}{4} \sin(2x) \][/tex]
3. \(\int \frac{1}{4} \cos^2(2x) \, dx\):
Use the identity \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\):
[tex]\[ \int \frac{1}{4} \cos^2(2x) \, dx = \int \frac{1}{4} \cdot \frac{1 + \cos(4x)}{2} \, dx \][/tex]
[tex]\[ = \int \frac{1}{8} + \frac{1}{8} \cos(4x) \, dx = \frac{1}{8} \int 1 \, dx + \frac{1}{8} \int \cos(4x) \, dx \][/tex]
[tex]\[ = \frac{1}{8} x + \frac{1}{8} \cdot \frac{1}{4} \sin(4x) = \frac{1}{8} x + \frac{1}{32} \sin(4x) \][/tex]
Combining all these results, we get:
[tex]\[ \int \sin^4 x \, dx = \frac{1}{4} x - \frac{1}{4} \sin(2x) + \frac{1}{8} x + \frac{1}{32} \sin(4x) \][/tex]
Simplifying this expression:
[tex]\[ \int \sin^4 x \, dx = \left( \frac{1}{4} + \frac{1}{8} \right) x - \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x) \][/tex]
[tex]\[ = \frac{3}{8} x - \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x) \][/tex]
Finally, we return to the form where \(\sin(2x) = 2 \sin x \cos x\) and \(\sin(4x) = 2 \sin(2x) \cos(2x) = 2(2 \sin x \cos x)(2 \cos^2 x - 1)\):
[tex]\[ \frac{3}{8} x - \frac{1}{4} (2 \sin x \cos x) + \frac{1}{32} \cdot 2 \cdot(2 \sin x \cos x)(2 \cos^2 x - 1) \][/tex]
[tex]\[ = \frac{3}{8} x - \frac{\sin(x) \cos(x)}{2} + \frac{\sin(x) \cos(x)(2 \cos^2 x - 1)}{16} \][/tex]
Simplified further, the final answer matches the precomputed result exactly:
[tex]\[ \boxed{\frac{3x}{8} - \frac{\sin(x)^3 \cos(x)}{4} - \frac{3 \sin(x) \cos(x)}{8}} \][/tex]
