Question

\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]

Answer 1

` ∫   {  sin  8x   }/{\sqrt{ 9 + sin^4  4x  }} `
\[ \Rightarrow \int\frac{2 \sin \left( 4x \right) \cdot \cos \left( 4 x \right)}{\sqrt{9 + \left( \sin^2 \left( 4x \right) \right)^2}}dx\]
\[\text{ let }\sin^2 \left( 4x \right) = t\]
\[ \Rightarrow 2 \text{ sin }\left( \text{ 4x } \right) \cdot \cos 4x \times \text{ 4 dx } = dt\]
\[ \Rightarrow 2 \text{ sin } \left( 4x \right) \cos \left( \text{ 4x }\right) dx = \frac{dt}{4}\]
\[Now, \int\frac{2 \text{ sin   }\left( 4x \right) \cdot \text{ cos }\left( 4 x \right)}{\sqrt{9 + \left( \sin^2 \left( 4x \right) \right)^2}}dx\]
\[ = \frac{1}{4}\int\frac{dt}{\sqrt{9 + t^2}}\]

\[ = \frac{1}{4}\int\frac{dt}{\sqrt{3^2 + t^2}}\]
\[ = \frac{1}{4} \text{ log }\left| t + \sqrt{3^2 + t^2} \right| + C\]
\[ = \frac{1}{4} \text{ log } \left| \sin^2 4x + \sqrt{9 + \sin^4 4x} \right| + C\]