Question

If $\int {\dfrac{{{2^x}}}{{\sqrt {1 - {4^x}} }}} dx = K{\sin ^{ - 1}}({2^x}) + C$, then $K$ is equal to
$\begin{align}
  A - \ln 2
  B - \dfrac{1}{2}\ln 2
  C - \dfrac{1}{2}
  D - \dfrac{1}{{\ln 2}}
\end{align} $

Answer 1

Hint: The chapter on integration has lots of formulae. Student should learn each and every formula to solve the problems. In this particular sum student should apply a substitution method to solve the numerical. This substitution should strike students' mind immediately on seeing the LHS and RHS .For such sums it is advisable to take up LHS and ignore RHS for the time being. Then the student should then start the sum by substituting the numerator and then solving. After substitution, he should take the derivative in order to bring the sum where we can apply the formula of integration. After this last step is comparing LHS and RHS, to find the value of $K$.

Complete step-by-step answer:
Let us first consider only LHS.
Consider $I = \int {\dfrac{{{2^x}}}{{\sqrt {1 - {4^x}} }}dx} $
Let us substitute ${2^x} = a..........(1)$
Now, taking derivative on both sides
${2^x}\log 2dx = da..........(2)$
We have applied the formula $\dfrac{d}{{dx}}({a^x}) = {a^x}\log a$ on LHS.
We can substitute the value of ${2^x}dx$ from Equation$2$ in the given sum. After substitution we get following step
$I = \int {\dfrac{{\dfrac{1}{{\log 2}}}}{{\sqrt {1 - {a^2}} }}da} ........(3)$
Since $\dfrac{1}{{\log 2}}$ is a constant , we can keep it outside the integration sign in order to remove complexity.
Now,
$I = \dfrac{1}{{\log 2}}\int {\dfrac{1}{{\sqrt {1 - {a^2}} }}da} ........(4)$
Now from the Equation $4$, we can figure out that the terms inside the integration sign form a direct formula . The value of this integration is ${\sin ^{ - 1}}a + C$.
Thus we can apply integration to Equation $4$ , and final answer before re-substitution would be
\[I = \dfrac{1}{{\log 2}}{\sin ^{ - 1}}a + C........(5)\]
Now we can re-substitute the original value of $a$ in Equation $5$. The final Expression is
\[I = \dfrac{1}{{\log 2}}{\sin ^{ - 1}}({2^x}) + C........(6)\]
Now comparing RHS and Equation $6$we can get the value of $K$.
Thus the value of $K$ is $\dfrac{1}{{\log 2}}$

The Answer to this problem is option $D - \dfrac{1}{{\ln 2}}$.

Note: There is nothing difficult in such problems. Only thing the student has to keep in mind is the substitution. If the students go wrong in substituting the entire sum would go for a toss. Thus it is advisable that students should properly understand the sum before progressing. Also it is vital to memorise each and every formula of Integration thoroughly he/she would not be able to solve a single sum where Application of integration is to be used.