question_answer

Evaluate \[\int{{{\sin }^{4}}x}{{\cos }^{4}}x\,\,dx.\]

Answer:

Let \[l=\int{{{\sin }^{4}}x{{\cos }^{4}}xdx}\] \[=\frac{1}{16}\int{16{{\sin }^{4}}x\cdot {{\cos }^{4}}xdx}\] \[=\frac{1}{16}\int{{{(2\,sin\,x\cdot \cos \,x)}^{4}}dx}\] \[=\frac{1}{16}\int{{{\sin }^{4}}2}xdx\]                      \[[\because 2sin\theta \cdot cos\theta =sin2\theta ]\] \[=\frac{1}{16}\int{{{({{\sin }^{2}}2x)}^{2}}dx}\] \[=\frac{1}{16}\int{{{\left( \frac{1-\cos 4x}{2} \right)}^{2}}dx}\] \[\left[ \because {{\sin }^{2}}\theta =\frac{1-\cos 2\theta }{2} \right]\]\[=\frac{1}{64}\int{(1+{{\cos }^{2}}4x-2\cos 4x)\,}dx\] \[=\frac{1}{64}\int{\left( 1+\frac{1+\cos 8x}{2}-2\cos 4x \right)dx}\] \[\left[ \because {{\cos }^{2}}\theta =\frac{1+\cos 2\theta }{2} \right]\] \[=\frac{1}{64}\int{\frac{2+1+\cos 8x-4\cos 4x}{2}\,}dx\] \[=\frac{1}{128}\int{(3+\cos 8x-4\cos 4x)\,dx}\] \[=\frac{1}{128}\left[ 3x-\sin 4x+\frac{1}{8}\sin x \right]+C\]