Evaluate the integral: I = integral sin^4 x dx
Evaluate the integral: $$ I = \int \sin^4 x\,dx. $$
Explanation of Method
Applying the power-reduction identity twice — once to write $\sin^4x=(\sin^2x)^2$ in terms of $\cos2x$, then again to reduce the resulting $\cos^2(2x)$ term — expresses $\sin^4x$ as a sum of a constant, a $\cos2x$ term, and a $\cos4x$ term, each easy to integrate.
- Rewrite $\sin^4x$ as $(\sin^2x)^2$ and apply the power-reduction identity.
- Reduce the resulting $\cos^2(2x)$ term with the identity again.
- Simplify into a sum of three terms.
- Integrate term by term.
Step 1 · First power reduction
Using $\sin^2x=\dfrac{1-\cos2x}{2}$:
$$ \sin^4x = \left(\sin^2x\right)^2 = \left(\frac{1-\cos2x}{2}\right)^2 = \frac{1-2\cos2x+\cos^2(2x)}{4}. $$
Step 2 · Second power reduction
Apply the identity again to $\cos^2(2x)=\dfrac{1+\cos4x}{2}$:
$$ \sin^4x = \frac{1-2\cos2x+\dfrac{1+\cos4x}{2}}{4}. $$
Step 3 · Simplify
Combining terms over a common denominator:
$$ \sin^4x = \frac{3}{8} - \frac12\cos2x + \frac18\cos4x. $$
Step 4 · Integrate term by term
$$ I = \int \left(\frac38 - \frac12\cos2x + \frac18\cos4x\right) dx = \frac38x - \frac14\sin2x + \frac{1}{32}\sin4x + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $3/8$ | $3x/8$ |
| $-\tfrac12\cos2x$ | $-\tfrac14\sin2x$ |
| $\tfrac18\cos4x$ | $\tfrac1{32}\sin4x$ |
Additional Notes
- The integrand is defined for all real $x$.
- Higher even powers of sine or cosine always reduce to a sum of $\cos(2kx)$ terms via repeated application of the power-reduction identity — the number of applications needed equals half the power.
- The result can be checked by differentiating $\tfrac38x-\tfrac14\sin2x+\tfrac1{32}\sin4x$ and confirming it returns $\sin^4x$.
Result
The solution to the integral is: