cos4 2x = (cos2 2x)2
\(=(\frac{1+cos4x}{2})^2\)
=\(\frac14\)[1+cos2 4x+2cos 4x]
\(= (\frac{1}{4}) [1+(\frac{1+cos8x}{2})+2cos4x]\)
\(= (\frac{1}{4}) [1+\frac{1}{2}+\frac{cos8x}{2}+2cos4x]\)
\(= (\frac{1}{4}) [ \frac{3}{2}+\frac{cos8x}{2}+2cos4x]\)
∴ ∫cos4 2x dx = ∫\((\frac{3}{8}+\frac{cos8x}{8}+\frac{cos4x}{2})dx\)
\(=\frac{3}{8}x+\frac{sin8x}{64}+\frac{sin4x}{8}+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
Observe the given sequence of nitrogenous bases on a DNA fragment and answer the following questions: 
(a) Name the restriction enzyme which can recognise the DNA sequence.
(b) Write the sequence after restriction enzyme cut the palindrome.
(c) Why are the ends generated after digestion called as ‘Sticky Ends’?
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C