Solveeit Logo
Login

Question

Question: If \(\int {\dfrac{{{2^x}}}{{\sqrt {1 - {4^x}} }}} dx = K{\sin ^{ - 1}}({2^x}) + C\), then \(K\) is e...

If ∫2x1−4xdx=Ksin⁡−1(2x)+C\int {\dfrac{{{2^x}}}{{\sqrt {1 - {4^x}} }}} dx = K{\sin ^{ - 1}}({2^x}) + C, then KK is equal to
A−ln⁡2B−12ln⁡2C−12D−1ln⁡2\begin{aligned} A - \ln 2 B - \dfrac{1}{2}\ln 2 C - \dfrac{1}{2} D - \dfrac{1}{{\ln 2}} \end{aligned}

Explanation

Solution

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 KK.

Complete step-by-step answer:
Let us first consider only LHS.
Consider I=∫2x1−4xdxI = \int {\dfrac{{{2^x}}}{{\sqrt {1 - {4^x}} }}dx}
Let us substitute 2x=a..........(1){2^x} = a..........(1)
Now, taking derivative on both sides
2xlog⁡2dx=da..........(2){2^x}\log 2dx = da..........(2)
We have applied the formula ddx(ax)=axlog⁡a\dfrac{d}{{dx}}({a^x}) = {a^x}\log a on LHS.
We can substitute the value of 2xdx{2^x}dx from Equation22 in the given sum. After substitution we get following step
I=∫1log⁡21−a2da........(3)I = \int {\dfrac{{\dfrac{1}{{\log 2}}}}{{\sqrt {1 - {a^2}} }}da} ........(3)
Since 1log⁡2\dfrac{1}{{\log 2}} is a constant , we can keep it outside the integration sign in order to remove complexity.
Now,
I=1log⁡2∫11−a2da........(4)I = \dfrac{1}{{\log 2}}\int {\dfrac{1}{{\sqrt {1 - {a^2}} }}da} ........(4)
Now from the Equation 44, we can figure out that the terms inside the integration sign form a direct formula . The value of this integration is sin⁡−1a+C{\sin ^{ - 1}}a + C.
Thus we can apply integration to Equation 44 , and final answer before re-substitution would be
I=1log⁡2sin⁡−1a+C........(5)I = \dfrac{1}{{\log 2}}{\sin ^{ - 1}}a + C........(5)
Now we can re-substitute the original value of aa in Equation 55. The final Expression is
I=1log⁡2sin⁡−1(2x)+C........(6)I = \dfrac{1}{{\log 2}}{\sin ^{ - 1}}({2^x}) + C........(6)
Now comparing RHS and Equation 66we can get the value of KK.
Thus the value of KK is 1log⁡2\dfrac{1}{{\log 2}}

The Answer to this problem is option D−1ln⁡2D - \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.