Integral Calculus
86277
\(\int \sin ^{4} x d x\) is equal to
1 \(\frac{1}{8}\left[3 x+\frac{\sin 4 x}{4}-\frac{4 \sin 2 x}{2}\right]+c\)
2 \(\frac{1}{8}\left[3 x+\frac{\sin 4 x}{4}+\frac{4 \sin 2 x}{2}\right]+c\)
3 \(\frac{1}{4}\left[3 x+\frac{\sin 4 x}{4}-\frac{4 \cos 2 x}{2}\right]+c\)
4 \(\frac{1}{4}\left[3 x+\frac{\sin 4 x}{4}+\frac{4 \cos 4 x}{2}\right]+c\)
Explanation:
Exp : (A) : Given
\(I=\int \sin ^4 x d x\)
\(=\int\left(\frac{1-\cos 2 x}{2}\right)^2 d x\)
\(=\frac{1}{4} \int\left(1+\cos ^2 2 x-2 \cos 2 x\right) d x\)
\(=\frac{1}{4} \int\left[1+\frac{1+\cos 4 x}{2}-2 \cos 2 x\right] d x\)
\(=\frac{1}{4} \int d x+\frac{1}{8} \int(1+\cos 4 x) d x-\frac{2}{4} \int \cos 2 x d x\)
\(=\frac{3}{8} \int d x+\frac{1}{8} \int \cos 4 x d x-\frac{1}{2} \int \cos 2 x d x\)
\(=\frac{3}{8} x+\frac{1}{32} \sin 4 x-\frac{1}{4} \sin 2 x+C\)
\(=\frac{1}{8}\left[3 x+\frac{\sin 4 x}{4}-\frac{4 \sin 2 x}{2}\right]+C\)